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


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


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


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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Just compare your series with the $p$-series $\sum\frac1{n^3}$. –  Brian M. Scott Nov 9 '12 at 18:41
This series is telescopic. –  P.. Nov 9 '12 at 19:24

4 Answers 4

up vote 6 down vote accepted

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


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:


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


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We haven't discussed integrals yet in the context of limits and sequences and series so your hint does not help that much - so any easier way to show that? Furthermore I understand that I can show that the series will converge by your mentioned approximation, however I still don't know how to get the limit in the second step. –  Christian Ivicevic Nov 9 '12 at 18:52
@Christian: Just observe that the Riemann sum $\sum_{k=1}^n\frac1{k^3}$ clearly exceeds $\int_1^{n+1}\frac{dx}{x^3}$, since $\frac1{x^3}$ is decreasing. Then $$\sum_{k\ge 1}\frac1{k^3}=\lim_{n\to\infty}\sum_{k=1}^n\frac1{k^3}\ge\lim_{n\to\infty}\int_1‌​^{n+1}\frac{dx}{x^3}=\int_1^\infty\frac{dx}{x^3}\;.$$ I’m not sure what you mean by the second step. The integral diverges, so the series diverges. –  Brian M. Scott Nov 9 '12 at 19:01
@amWhy: I included the integral argument only in case he didn’t already know the $p$-series test. But if it’s a fairly rigorous course, and he’s not seen the integral test, there’s a fair chance that he’s not seen the $p$-series test, either. –  Brian M. Scott Nov 9 '12 at 19:04
@BrianM.Scott No criticism intended (note the +1)! I encountered the p-series in Calculus II and again when first studying Baby Rudin neither of which appealed to integral. (In short, it was presented as part of one's "toolkit" for testing convergence/divergence of series, together with ratio test, root test, etc.) –  amWhy Nov 9 '12 at 19:07
@BrianM.Scott: Looking up in my textbook the $p$-series is prooved (in a part we did not work through yet) with approximations $1/n^k\leq 1/n^2\leq 2/(n(n+1))$ for $k>1,k\in\mathbb{N}$ and the last term converges because the telescopic series converges. You claim, that my series I have to work with does not converge, however WolframAlpha claims that it converges to $1/4$ - am I missing something? –  Christian Ivicevic Nov 9 '12 at 19:08

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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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 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.

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