Find $ sum_{n=1}^{infty} frac{1}{n(n+2)} $












2














I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
begin{equation}
sum_{n=1}^{infty} frac{1}{n(n+2)}
end{equation}
First I set $ frac{1}{n(n+2)}= frac{A}{n}+frac{B}{n+2}$ and solve to get $A=frac{1}{2}$ and $B=-frac{1}{2}$.
So I'm finding $sum_{n=1}^{infty}(frac{1}{2n}-frac{1}{2n+4})$
and proceed with the partial sum expansion.
At the end I am left with $S_N=frac{1}{2}+frac{1}{4}-frac{1}{2N+4}$ by cancelling terms in between and then $lim_{N to infty}S_N=frac{3}{4}$.



Am I doing it correctly? Thank you for your time.






real-analysis sequences-and-series







share|cite|improve this question




















  • 2




    yep
    – tilper
    yesterday










  • Thanks @tilper.
    – Allorja
    yesterday






  • 1




    Check again on your partial sums. I think you missed another negative term (though it tends to zero).
    – JavaMan
    yesterday










  • Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though.
    – Allorja
    yesterday
















2














I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
begin{equation}
sum_{n=1}^{infty} frac{1}{n(n+2)}
end{equation}
First I set $ frac{1}{n(n+2)}= frac{A}{n}+frac{B}{n+2}$ and solve to get $A=frac{1}{2}$ and $B=-frac{1}{2}$.
So I'm finding $sum_{n=1}^{infty}(frac{1}{2n}-frac{1}{2n+4})$
and proceed with the partial sum expansion.
At the end I am left with $S_N=frac{1}{2}+frac{1}{4}-frac{1}{2N+4}$ by cancelling terms in between and then $lim_{N to infty}S_N=frac{3}{4}$.



Am I doing it correctly? Thank you for your time.






real-analysis sequences-and-series







share|cite|improve this question




















  • 2




    yep
    – tilper
    yesterday










  • Thanks @tilper.
    – Allorja
    yesterday






  • 1




    Check again on your partial sums. I think you missed another negative term (though it tends to zero).
    – JavaMan
    yesterday










  • Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though.
    – Allorja
    yesterday














2












2








2







I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
begin{equation}
sum_{n=1}^{infty} frac{1}{n(n+2)}
end{equation}
First I set $ frac{1}{n(n+2)}= frac{A}{n}+frac{B}{n+2}$ and solve to get $A=frac{1}{2}$ and $B=-frac{1}{2}$.
So I'm finding $sum_{n=1}^{infty}(frac{1}{2n}-frac{1}{2n+4})$
and proceed with the partial sum expansion.
At the end I am left with $S_N=frac{1}{2}+frac{1}{4}-frac{1}{2N+4}$ by cancelling terms in between and then $lim_{N to infty}S_N=frac{3}{4}$.



Am I doing it correctly? Thank you for your time.






real-analysis sequences-and-series







share|cite|improve this question















I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
begin{equation}
sum_{n=1}^{infty} frac{1}{n(n+2)}
end{equation}
First I set $ frac{1}{n(n+2)}= frac{A}{n}+frac{B}{n+2}$ and solve to get $A=frac{1}{2}$ and $B=-frac{1}{2}$.
So I'm finding $sum_{n=1}^{infty}(frac{1}{2n}-frac{1}{2n+4})$
and proceed with the partial sum expansion.
At the end I am left with $S_N=frac{1}{2}+frac{1}{4}-frac{1}{2N+4}$ by cancelling terms in between and then $lim_{N to infty}S_N=frac{3}{4}$.



Am I doing it correctly? Thank you for your time.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Allorja

479




479








  • 2




    yep
    – tilper
    yesterday










  • Thanks @tilper.
    – Allorja
    yesterday






  • 1




    Check again on your partial sums. I think you missed another negative term (though it tends to zero).
    – JavaMan
    yesterday










  • Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though.
    – Allorja
    yesterday














  • 2




    yep
    – tilper
    yesterday










  • Thanks @tilper.
    – Allorja
    yesterday






  • 1




    Check again on your partial sums. I think you missed another negative term (though it tends to zero).
    – JavaMan
    yesterday










  • Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though.
    – Allorja
    yesterday








2




2




yep
– tilper
yesterday




yep
– tilper
yesterday












Thanks @tilper.
– Allorja
yesterday




Thanks @tilper.
– Allorja
yesterday




1




1




Check again on your partial sums. I think you missed another negative term (though it tends to zero).
– JavaMan
yesterday




Check again on your partial sums. I think you missed another negative term (though it tends to zero).
– JavaMan
yesterday












Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though.
– Allorja
yesterday




Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though.
– Allorja
yesterday










2 Answers
2






active

oldest

votes


















5














$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$



$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$



$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$






share|cite|improve this answer

















  • 1




    Sorry but did you read the question? (The same would go for four upvoters...)
    – Did
    yesterday





















2














In case you're interested, you can also compute this series using a certain representation of the digamma function $psi(z+1)=-gamma+sum_{nge 1} frac{z}{n(n+z)}$. We are then looking for $frac{psi(3)+gamma}{2}$. Using the following integral representation of the digamma function,
$$psi(s+1)=-gamma+int_0^1 frac{1-x^s}{1-x},dx$$
we find that $psi(3)=-gamma+frac{3}{2}$.
Your sum is then
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$






share|cite|improve this answer























  • Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
    – Allorja
    yesterday











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$



$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$



$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$






share|cite|improve this answer

















  • 1




    Sorry but did you read the question? (The same would go for four upvoters...)
    – Did
    yesterday


















5














$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$



$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$



$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$






share|cite|improve this answer

















  • 1




    Sorry but did you read the question? (The same would go for four upvoters...)
    – Did
    yesterday
















5












5








5






$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$



$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$



$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$






share|cite|improve this answer












$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$



$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$



$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









D Tiwari

5,3802630




5,3802630








  • 1




    Sorry but did you read the question? (The same would go for four upvoters...)
    – Did
    yesterday
















  • 1




    Sorry but did you read the question? (The same would go for four upvoters...)
    – Did
    yesterday










1




1




Sorry but did you read the question? (The same would go for four upvoters...)
– Did
yesterday






Sorry but did you read the question? (The same would go for four upvoters...)
– Did
yesterday













2














In case you're interested, you can also compute this series using a certain representation of the digamma function $psi(z+1)=-gamma+sum_{nge 1} frac{z}{n(n+z)}$. We are then looking for $frac{psi(3)+gamma}{2}$. Using the following integral representation of the digamma function,
$$psi(s+1)=-gamma+int_0^1 frac{1-x^s}{1-x},dx$$
we find that $psi(3)=-gamma+frac{3}{2}$.
Your sum is then
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$






share|cite|improve this answer























  • Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
    – Allorja
    yesterday
















2














In case you're interested, you can also compute this series using a certain representation of the digamma function $psi(z+1)=-gamma+sum_{nge 1} frac{z}{n(n+z)}$. We are then looking for $frac{psi(3)+gamma}{2}$. Using the following integral representation of the digamma function,
$$psi(s+1)=-gamma+int_0^1 frac{1-x^s}{1-x},dx$$
we find that $psi(3)=-gamma+frac{3}{2}$.
Your sum is then
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$






share|cite|improve this answer























  • Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
    – Allorja
    yesterday














2












2








2






In case you're interested, you can also compute this series using a certain representation of the digamma function $psi(z+1)=-gamma+sum_{nge 1} frac{z}{n(n+z)}$. We are then looking for $frac{psi(3)+gamma}{2}$. Using the following integral representation of the digamma function,
$$psi(s+1)=-gamma+int_0^1 frac{1-x^s}{1-x},dx$$
we find that $psi(3)=-gamma+frac{3}{2}$.
Your sum is then
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$






share|cite|improve this answer














In case you're interested, you can also compute this series using a certain representation of the digamma function $psi(z+1)=-gamma+sum_{nge 1} frac{z}{n(n+z)}$. We are then looking for $frac{psi(3)+gamma}{2}$. Using the following integral representation of the digamma function,
$$psi(s+1)=-gamma+int_0^1 frac{1-x^s}{1-x},dx$$
we find that $psi(3)=-gamma+frac{3}{2}$.
Your sum is then
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Zachary

2,2971213




2,2971213












  • Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
    – Allorja
    yesterday


















  • Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
    – Allorja
    yesterday
















Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
– Allorja
yesterday




Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though.
– Allorja
yesterday


















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