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Does the following series converge? If it does, determine the appropriate limit.

$\sum\limits_{k=1}^\infty\left(\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}\right)$

The only thing i noticed so far is the occurence od the telescopic series via a transformation:

$\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}=\frac{1}{k(k+1)(k+2)}$

The ratio test delivers the result $k/(k+3)$ which renders it unhelpful, so I have to try something else. Now I have been thinking about finding an explicit expression for the partial sums

$\sum\limits_{k=1}^N\left(\frac{1}{k(k+1)(k+2)}\right)$

however I neither know, how do so nor do I know whether it suffices to show, that the partial sums will converge for $N\to\infty$ to conclude that the whole series will have a limit.

I need help on how to determine the the explicit expression of the partial sums and would like to know some good suggestions on what to do else.

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    This series is telescopic.2012-11-09

4 Answers 4

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You’re making it harder than necessary: for each $k\in\Bbb Z^+$ you have

$\frac1{k(k+1)(k+2)}<\frac1{k^3}\;;$

Now just use the ordinary comparison test.

I’m assuming that you’ve already shown that $\sum_{k\ge 1}\frac1{k^p}$ converges for all $p>1$. If not, note that

$\sum_{k\ge 1}\frac1{k^p}\ge\int_1^\infty\frac{dx}{x^p}\;,$

which diverges for $p>1$.

Added: In view of the comments, I’ll suggest another approach, a variation on partial fractions. In order to get something that might telescope, you want ideally two terms with denominators that are offset by $1$, so try this decomposition:

$\frac1{k(k+1)(k+2)}=\frac{A}{k(k+1)}+\frac{B}{(k+1)(k+2)}\;;$

clearly $B=-A$ and $A=\frac12$, so $\frac1{k(k+1)(k+2)}=\frac12\left(\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right)\;,$ and your partial sum is

$\frac12\sum_{k=1}^N\left(\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right)\;.$

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    @Christian: Yes, I did; I’ll fix it in a moment. You’re welcome!2012-11-09
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Hint: $\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\left(\dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right).$
Now use the method of differences to find an explicit expression for the partial sums.

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You can bound each term from above by $k^{-3}$. Do you know that sum converges? If not, you can bound that from above by the integral of $x^{-3}$

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You may write it in the form $x_k=a_{k+2}-2a_{k+1}+a_k$ (you're almost there). Thus, if you write out the partial sums

$x_1+ \cdots + \cdots + x_n$

cancelling out the corresponding intermediate terms, you will see that since $a_n \to 0$, the sum clearly converges.

You're almost there.