Mixing
Find $ sum_{n=1}^{infty} frac{1}{n(n+2)} $
$begingroup$
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
asked Jan 3 at 14:01
AllorjaAllorja
699
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 14:01
AllorjaAllorja
699
$endgroup$
2
$begingroup$
yep
$endgroup$
– tilper
Jan 3 at 14:04
$begingroup$
Thanks @tilper.
$endgroup$
– Allorja
Jan 3 at 14:06
1
$begingroup$
Check again on your partial sums. I think you missed another negative term (though it tends to zero).
$endgroup$
– JavaMan
Jan 3 at 14:07
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 14:10
add a comment |
$begingroup$
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
asked Jan 3 at 14:01
AllorjaAllorja
699
$endgroup$
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
asked Jan 3 at 14:01
AllorjaAllorja
699
asked Jan 3 at 14:01
AllorjaAllorja
699
edited Jan 4 at 0:00
Allorja
asked Jan 3 at 14:01
AllorjaAllorja
699
asked Jan 3 at 14:01
AllorjaAllorja
699
asked Jan 3 at 14:01
AllorjaAllorja
699
699
2
$begingroup$
yep
$endgroup$
– tilper
Jan 3 at 14:04
$begingroup$
Thanks @tilper.
$endgroup$
– Allorja
Jan 3 at 14:06
1
$begingroup$
Check again on your partial sums. I think you missed another negative term (though it tends to zero).
$endgroup$
– JavaMan
Jan 3 at 14:07
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 14:10
add a comment |
2
$begingroup$
yep
$endgroup$
– tilper
Jan 3 at 14:04
$begingroup$
Thanks @tilper.
$endgroup$
– Allorja
Jan 3 at 14:06
1
$begingroup$
Check again on your partial sums. I think you missed another negative term (though it tends to zero).
$endgroup$
– JavaMan
Jan 3 at 14:07
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 14:10
2
2
$begingroup$
yep
$endgroup$
– tilper
Jan 3 at 14:04
$begingroup$
yep
$endgroup$
– tilper
Jan 3 at 14:04
$begingroup$
Thanks @tilper.
$endgroup$
– Allorja
Jan 3 at 14:06
$begingroup$
Thanks @tilper.
$endgroup$
– Allorja
Jan 3 at 14:06
1
1
$begingroup$
Check again on your partial sums. I think you missed another negative term (though it tends to zero).
$endgroup$
– JavaMan
Jan 3 at 14:07
$begingroup$
Check again on your partial sums. I think you missed another negative term (though it tends to zero).
$endgroup$
– JavaMan
Jan 3 at 14:07
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 14:10
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 14:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$
$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$
$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$
answered Jan 3 at 14:08
DXTDXT
5,4462630
$endgroup$
1
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
add a comment |
$begingroup$
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
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
$endgroup$
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$
$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$
$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$
answered Jan 3 at 14:08
DXTDXT
5,4462630
$endgroup$
1
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
add a comment |
$begingroup$
$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$
$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$
$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$
answered Jan 3 at 14:08
DXTDXT
5,4462630
$endgroup$
1
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
add a comment |
$begingroup$
$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$
$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$
$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$
answered Jan 3 at 14:08
DXTDXT
5,4462630
$endgroup$
$$sum^{infty}_{n=1}frac{1}{n(n+2)} = frac{1}{2}sum^{infty}_{n=1}bigg[frac{1}{n}-frac{1}{n+2}bigg]$$
$$ = frac{1}{2}sum^{infty}_{n=1}bigg[bigg(frac{1}{n}-frac{1}{n+1}bigg)+bigg(frac{1}{n+1}-frac{1}{n+2}bigg)bigg]$$
$$ = frac{1}{2}bigg[frac{1}{1}+frac{1}{2}bigg] = frac{3}{4}$$
answered Jan 3 at 14:08
DXTDXT
5,4462630
answered Jan 3 at 14:08
DXTDXT
5,4462630
answered Jan 3 at 14:08
DXTDXT
5,4462630
answered Jan 3 at 14:08
DXTDXT
5,4462630
5,4462630
1
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
add a comment |
1
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
1
1
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
$begingroup$
Sorry but did you read the question? (The same would go for four upvoters...)
$endgroup$
– Did
Jan 3 at 15:22
add a comment |
$begingroup$
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
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
$endgroup$
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
add a comment |
$begingroup$
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
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
$endgroup$
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
add a comment |
$begingroup$
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
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
$endgroup$
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
$$sum_{nge 1} frac{1}{n(n+2)}=frac{3}{4}$$
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
edited Jan 3 at 22:13
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
answered Jan 3 at 15:20
ZacharyZachary
2,3151214
2,3151214
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
add a comment |
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 19:23
add a comment |
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2
$begingroup$
yep
$endgroup$
– tilper
Jan 3 at 14:04
$begingroup$
Thanks @tilper.
$endgroup$
– Allorja
Jan 3 at 14:06
1
$begingroup$
Check again on your partial sums. I think you missed another negative term (though it tends to zero).
$endgroup$
– JavaMan
Jan 3 at 14:07
$begingroup$
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.
$endgroup$
– Allorja
Jan 3 at 14:10