Ratio test to find radius of convergence: $sumlimits_{n=0}^infty (n+1)(t-1)^{2n}$












3












$begingroup$


Use the ratio test on :



$$sumlimits_{n=0}^infty (n+1)(t-1)^{2n}$$



and find the radius of convergence.



I know to start the ratio test, we consider $frac{a_{n+1}}{a_n}$



So I let:



$$a_n=(n+1)(t-1)^{2n}$$



$$a_{n+1}=(n+2)(t-1)^{2n+2}$$



Thus,



$$frac{a_{n+1}}{a_n} = frac{(n+2)(t-1)^{2n+2}}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^{2n}(t-1)^2}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^2}{(n+1)}$$



Then I take the limit:



$$(t-1)^2 lim_{ntoinfty} frac{n+2}{n+1}=(t-1)^2$$



Thus, for this series:




  • Convergence if $(t-1)^2<1$

  • Divergence if $(t-1)^2>1$


So is the radius of convergence $1$?






calculus sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. But to be complete, not that $(t-1)^2<1$ implies $|t-1| < 1$ and then guarantees convergence $0<t<2.$ Convergence at the endpoints $t=0$ and $t=2$ will not occur for this particular series.
    $endgroup$
    – matt biesecker
    May 11 '15 at 0:29
















3












$begingroup$


Use the ratio test on :



$$sumlimits_{n=0}^infty (n+1)(t-1)^{2n}$$



and find the radius of convergence.



I know to start the ratio test, we consider $frac{a_{n+1}}{a_n}$



So I let:



$$a_n=(n+1)(t-1)^{2n}$$



$$a_{n+1}=(n+2)(t-1)^{2n+2}$$



Thus,



$$frac{a_{n+1}}{a_n} = frac{(n+2)(t-1)^{2n+2}}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^{2n}(t-1)^2}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^2}{(n+1)}$$



Then I take the limit:



$$(t-1)^2 lim_{ntoinfty} frac{n+2}{n+1}=(t-1)^2$$



Thus, for this series:




  • Convergence if $(t-1)^2<1$

  • Divergence if $(t-1)^2>1$


So is the radius of convergence $1$?






calculus sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. But to be complete, not that $(t-1)^2<1$ implies $|t-1| < 1$ and then guarantees convergence $0<t<2.$ Convergence at the endpoints $t=0$ and $t=2$ will not occur for this particular series.
    $endgroup$
    – matt biesecker
    May 11 '15 at 0:29














3












3








3





$begingroup$


Use the ratio test on :



$$sumlimits_{n=0}^infty (n+1)(t-1)^{2n}$$



and find the radius of convergence.



I know to start the ratio test, we consider $frac{a_{n+1}}{a_n}$



So I let:



$$a_n=(n+1)(t-1)^{2n}$$



$$a_{n+1}=(n+2)(t-1)^{2n+2}$$



Thus,



$$frac{a_{n+1}}{a_n} = frac{(n+2)(t-1)^{2n+2}}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^{2n}(t-1)^2}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^2}{(n+1)}$$



Then I take the limit:



$$(t-1)^2 lim_{ntoinfty} frac{n+2}{n+1}=(t-1)^2$$



Thus, for this series:




  • Convergence if $(t-1)^2<1$

  • Divergence if $(t-1)^2>1$


So is the radius of convergence $1$?






calculus sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$




Use the ratio test on :



$$sumlimits_{n=0}^infty (n+1)(t-1)^{2n}$$



and find the radius of convergence.



I know to start the ratio test, we consider $frac{a_{n+1}}{a_n}$



So I let:



$$a_n=(n+1)(t-1)^{2n}$$



$$a_{n+1}=(n+2)(t-1)^{2n+2}$$



Thus,



$$frac{a_{n+1}}{a_n} = frac{(n+2)(t-1)^{2n+2}}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^{2n}(t-1)^2}{(n+1)(t-1)^{2n}} = frac{(n+2)(t-1)^2}{(n+1)}$$



Then I take the limit:



$$(t-1)^2 lim_{ntoinfty} frac{n+2}{n+1}=(t-1)^2$$



Thus, for this series:




  • Convergence if $(t-1)^2<1$

  • Divergence if $(t-1)^2>1$


So is the radius of convergence $1$?






calculus sequences-and-series proof-verification convergence




calculus sequences-and-series proof-verification convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 7:31









Eevee Trainer

5,4691936




5,4691936










asked May 11 '15 at 0:21









ChilanieChilanie

174110




174110












  • $begingroup$
    Yes. But to be complete, not that $(t-1)^2<1$ implies $|t-1| < 1$ and then guarantees convergence $0<t<2.$ Convergence at the endpoints $t=0$ and $t=2$ will not occur for this particular series.
    $endgroup$
    – matt biesecker
    May 11 '15 at 0:29


















  • $begingroup$
    Yes. But to be complete, not that $(t-1)^2<1$ implies $|t-1| < 1$ and then guarantees convergence $0<t<2.$ Convergence at the endpoints $t=0$ and $t=2$ will not occur for this particular series.
    $endgroup$
    – matt biesecker
    May 11 '15 at 0:29
















$begingroup$
Yes. But to be complete, not that $(t-1)^2<1$ implies $|t-1| < 1$ and then guarantees convergence $0<t<2.$ Convergence at the endpoints $t=0$ and $t=2$ will not occur for this particular series.
$endgroup$
– matt biesecker
May 11 '15 at 0:29




$begingroup$
Yes. But to be complete, not that $(t-1)^2<1$ implies $|t-1| < 1$ and then guarantees convergence $0<t<2.$ Convergence at the endpoints $t=0$ and $t=2$ will not occur for this particular series.
$endgroup$
– matt biesecker
May 11 '15 at 0:29










1 Answer
1






active

oldest

votes


















1












$begingroup$

Your solution for the most part is correct, up until determining the radius of convergence. (Your solution is correct, but incomplete, and I suspect correct for the wrong reasons since you didn't explicitly solve for $t$.) We note: for convergence, we want



$$(t-1)^2 < 1$$



Taking the square root of both sides gives the following, depending on whether you take the negative or positive root:



$$t-1 < 1 ;;; text{and} ;;; t-1 > -1$$



An equivalent formulation of this statement is $|t-1|<1$. (You can see why this statement holds easily by solving $x^2 < 1$. From this, solving for $t$:



$$t < 2 ;;; text{and} ;;; t > 0$$



Thus, for all $t$ satisfying $0 < t < 2$, the series converges, and diverges otherwise. The width of this interval is $2$, thus giving a radius of convergence of $1$. If you wanted to go further into the details on the interval of convergence, you could also justify the non-inclusion of $t=0$ and $t=2$ fairly easily by plugging each into the original series, simplifying, and checking for convergence that way as well.



This would be starkly different if, say, we have $(3t-1)^2 < 1$ instead for this series, which would result in a radius of convergence of $1/3$. (I'll leave showing that as an exercise to the reader.) The key point here being to make sure your work is all justified and to see where it comes from. I could easily see a few number of ways you could just pick out $1$ as the radius of convergence from $(t-1)^2 < 1$ and, while having the right radius, you would be wrong if you applied it to similar problems.



In short, correct solution, just perhaps for not the right reasons, it's hard to say, and this thus should be an example of the importance of showing your work.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
    $endgroup$
    – DavidG
    Jan 8 at 4:46











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1 Answer
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1












$begingroup$

Your solution for the most part is correct, up until determining the radius of convergence. (Your solution is correct, but incomplete, and I suspect correct for the wrong reasons since you didn't explicitly solve for $t$.) We note: for convergence, we want



$$(t-1)^2 < 1$$



Taking the square root of both sides gives the following, depending on whether you take the negative or positive root:



$$t-1 < 1 ;;; text{and} ;;; t-1 > -1$$



An equivalent formulation of this statement is $|t-1|<1$. (You can see why this statement holds easily by solving $x^2 < 1$. From this, solving for $t$:



$$t < 2 ;;; text{and} ;;; t > 0$$



Thus, for all $t$ satisfying $0 < t < 2$, the series converges, and diverges otherwise. The width of this interval is $2$, thus giving a radius of convergence of $1$. If you wanted to go further into the details on the interval of convergence, you could also justify the non-inclusion of $t=0$ and $t=2$ fairly easily by plugging each into the original series, simplifying, and checking for convergence that way as well.



This would be starkly different if, say, we have $(3t-1)^2 < 1$ instead for this series, which would result in a radius of convergence of $1/3$. (I'll leave showing that as an exercise to the reader.) The key point here being to make sure your work is all justified and to see where it comes from. I could easily see a few number of ways you could just pick out $1$ as the radius of convergence from $(t-1)^2 < 1$ and, while having the right radius, you would be wrong if you applied it to similar problems.



In short, correct solution, just perhaps for not the right reasons, it's hard to say, and this thus should be an example of the importance of showing your work.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
    $endgroup$
    – DavidG
    Jan 8 at 4:46
















1












$begingroup$

Your solution for the most part is correct, up until determining the radius of convergence. (Your solution is correct, but incomplete, and I suspect correct for the wrong reasons since you didn't explicitly solve for $t$.) We note: for convergence, we want



$$(t-1)^2 < 1$$



Taking the square root of both sides gives the following, depending on whether you take the negative or positive root:



$$t-1 < 1 ;;; text{and} ;;; t-1 > -1$$



An equivalent formulation of this statement is $|t-1|<1$. (You can see why this statement holds easily by solving $x^2 < 1$. From this, solving for $t$:



$$t < 2 ;;; text{and} ;;; t > 0$$



Thus, for all $t$ satisfying $0 < t < 2$, the series converges, and diverges otherwise. The width of this interval is $2$, thus giving a radius of convergence of $1$. If you wanted to go further into the details on the interval of convergence, you could also justify the non-inclusion of $t=0$ and $t=2$ fairly easily by plugging each into the original series, simplifying, and checking for convergence that way as well.



This would be starkly different if, say, we have $(3t-1)^2 < 1$ instead for this series, which would result in a radius of convergence of $1/3$. (I'll leave showing that as an exercise to the reader.) The key point here being to make sure your work is all justified and to see where it comes from. I could easily see a few number of ways you could just pick out $1$ as the radius of convergence from $(t-1)^2 < 1$ and, while having the right radius, you would be wrong if you applied it to similar problems.



In short, correct solution, just perhaps for not the right reasons, it's hard to say, and this thus should be an example of the importance of showing your work.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
    $endgroup$
    – DavidG
    Jan 8 at 4:46














1












1








1





$begingroup$

Your solution for the most part is correct, up until determining the radius of convergence. (Your solution is correct, but incomplete, and I suspect correct for the wrong reasons since you didn't explicitly solve for $t$.) We note: for convergence, we want



$$(t-1)^2 < 1$$



Taking the square root of both sides gives the following, depending on whether you take the negative or positive root:



$$t-1 < 1 ;;; text{and} ;;; t-1 > -1$$



An equivalent formulation of this statement is $|t-1|<1$. (You can see why this statement holds easily by solving $x^2 < 1$. From this, solving for $t$:



$$t < 2 ;;; text{and} ;;; t > 0$$



Thus, for all $t$ satisfying $0 < t < 2$, the series converges, and diverges otherwise. The width of this interval is $2$, thus giving a radius of convergence of $1$. If you wanted to go further into the details on the interval of convergence, you could also justify the non-inclusion of $t=0$ and $t=2$ fairly easily by plugging each into the original series, simplifying, and checking for convergence that way as well.



This would be starkly different if, say, we have $(3t-1)^2 < 1$ instead for this series, which would result in a radius of convergence of $1/3$. (I'll leave showing that as an exercise to the reader.) The key point here being to make sure your work is all justified and to see where it comes from. I could easily see a few number of ways you could just pick out $1$ as the radius of convergence from $(t-1)^2 < 1$ and, while having the right radius, you would be wrong if you applied it to similar problems.



In short, correct solution, just perhaps for not the right reasons, it's hard to say, and this thus should be an example of the importance of showing your work.






share|cite|improve this answer









$endgroup$



Your solution for the most part is correct, up until determining the radius of convergence. (Your solution is correct, but incomplete, and I suspect correct for the wrong reasons since you didn't explicitly solve for $t$.) We note: for convergence, we want



$$(t-1)^2 < 1$$



Taking the square root of both sides gives the following, depending on whether you take the negative or positive root:



$$t-1 < 1 ;;; text{and} ;;; t-1 > -1$$



An equivalent formulation of this statement is $|t-1|<1$. (You can see why this statement holds easily by solving $x^2 < 1$. From this, solving for $t$:



$$t < 2 ;;; text{and} ;;; t > 0$$



Thus, for all $t$ satisfying $0 < t < 2$, the series converges, and diverges otherwise. The width of this interval is $2$, thus giving a radius of convergence of $1$. If you wanted to go further into the details on the interval of convergence, you could also justify the non-inclusion of $t=0$ and $t=2$ fairly easily by plugging each into the original series, simplifying, and checking for convergence that way as well.



This would be starkly different if, say, we have $(3t-1)^2 < 1$ instead for this series, which would result in a radius of convergence of $1/3$. (I'll leave showing that as an exercise to the reader.) The key point here being to make sure your work is all justified and to see where it comes from. I could easily see a few number of ways you could just pick out $1$ as the radius of convergence from $(t-1)^2 < 1$ and, while having the right radius, you would be wrong if you applied it to similar problems.



In short, correct solution, just perhaps for not the right reasons, it's hard to say, and this thus should be an example of the importance of showing your work.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 7:29









Eevee TrainerEevee Trainer

5,4691936




5,4691936












  • $begingroup$
    There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
    $endgroup$
    – DavidG
    Jan 8 at 4:46


















  • $begingroup$
    There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
    $endgroup$
    – DavidG
    Jan 8 at 4:46
















$begingroup$
There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
$endgroup$
– DavidG
Jan 8 at 4:46




$begingroup$
There is detail missing here. When applying a function to an inequality, it must first be established that the function is monotonic on the interval. If it isn't, you must break it apart into the different intervals of monotonic behaviour. Within each monotonic sets you then need to identify whether the function is increasing or decreasing as in the case of decreasing the inequality sign is reversed. Here taking the positive square root on the interval, we have a function that is monotonic and increasing and so, it is allowable here.
$endgroup$
– DavidG
Jan 8 at 4:46


















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