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I want to find the following series.

$$\sum_{n\ge1} \frac{1}{4n^3-n}$$

So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$

Next, I consider the geometric series $$\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-x^2}+\frac{x^2}{1-x^2}-\frac{1}{1-x}\tag{1}$$ where ${-1\lt x\lt1}$.

Then, I integrate both side of (1) from $0$ to $1$:

$$\sum_{n\ge1}\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n}=\int_0^1 \frac{x^2-x}{1-x^2}dx$$

Finally, I get the value that is equal to $ln(2)-1$ by improper integral but it's less than $0$. What did I do wrong? All help is appreciated.

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    it seems that this sum is not so easy to compute – 2017-01-20
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    This series is telescopic? I can't do it with telescopic. – 2017-01-20
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    @XTL Telescope it how, exactly? – 2017-01-20
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    To be clear: are you asking specifically for the method your\ are suggesting (once fixed/massaged), or are you open to other techniques? (e.g., I have one involving partial sums and the Harmonic series, yielding $2\ln2 -1$). – 2017-01-20
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    (I would say, without checking too carefully, that the issue in your step most likely resides in the "I integrate both sides of (1) from 0 to 1." You integrate the LHS termwise -- why can you do that? The interval of convergence is $1$, so it is not obvious you can swap $\sum$ and $\int_0^1$. And, as it turns out, worse than "not obvious" it seems to be wrong) – 2017-01-20
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    please, show all your techniques. @ClementC. – 2017-01-20
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    Your integration in $\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}$ isn't valid. First item not defined integration. – 2017-01-20
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    How? @MyGlasses. I have already integrated from 0 to 1. – 2017-01-20
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    @MoNtiDeaDMoonDogs See below. – 2017-01-20
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    $$ \begin{align} &\frac{1}{4n^3-n}=\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{\color{red}{2}}{\color{red}{2}\,n}\\ &\sum_{n=1}^{\infty}\left[x^{2n-2}+x^{2n}-\color{red}{2}\,x^{\color{red}{2}n-1}\right]=\frac{1}{1-x^2}+\frac{x^2}{1-x^2}-\frac{2\,x}{1-x^2}=\frac{(1-x)^2}{1-x^2}=\frac{1-x}{1+x}\\ &\sum_{n=1}^{\infty}\frac{1}{4\,n^3-n}=\int_0^1\frac{1-x}{1+x}\,dx=\color{red}{2\log{2}-1} \end{align} $$ – 2017-01-20
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    What is different between $\frac{1}{n}$ and $\frac{2}{2n}$ for calculation? Why do you did like this?.@HazemOrabi – 2017-01-20
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    @MoNtiDeaDMoonDogs It changes the behavior of the series: $\sum_n x^{n-1}$ does not have the same behavior near $1$ as $\sum_n x^{2n-1}$ does. This is a neat trick, a bit like magic; but wouldn't be seen as a rigorous proof (at least by me as a TA) unless you do explain *why* this swapping series/integral is fine, while the previous one is not. – 2017-01-20
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    $$ \begin{align} \frac{1}{4n^3-n} &=\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n} \\ &=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{\color{red}{2}}{\color{red}{2}\,n}\\ &=\left[\frac{1}{2n-1}-\frac{\color{red}{1}}{\color{red}{2}\,n}\right] + \left[\frac{1}{2n+1}-\frac{\color{red}{1}}{\color{red}{2}\,n}\right] \end{align} $$ – 2017-01-20
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2 Answers 2

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Following a comment of mine above:

  • without checking too carefully, the issue in your step most likely resides in the "I integrate both sides of (1) from 0 to 1." You then integrate the LHS termwise -- why can you do that? The interval of convergence of your power series is $1$, so it is not obvious you can swap $\sum_n$ and $\int_0^1$. And, as it turns out, worse than "not obvious" it appears to be wrong.

  • Now, another approach to compute the sum: starting with your $(1)$, we write, for $N\geq 1$, $$\begin{align} \sum_{n=1}^N a_n &= \sum_{n=1}^N \frac{1}{2n+1}+\sum_{n=1}^N \frac{1}{2n-1}-\sum_{n=1}^N \frac{1}{n}\\ &= \sum_{n=1}^N \left(\frac{1}{2n+1}+\frac{1}{2n}\right)+\sum_{n=1}^N \left(\frac{1}{2n-1}+\frac{1}{2n}\right)-2\sum_{n=1}^N \frac{1}{n} \\ &= \sum_{n=2}^{2N+1} \frac{1}{n}+\sum_{n=1}^{2N} \frac{1}{n}-2\sum_{n=1}^N \frac{1}{n} \\ &= H_{2N+1}-1+H_{2N}-2H_N \\ &= \ln(2N+1)+\ln(2N)-2\ln N - 1 +o(1)\\ &= 2\ln 2 - 1 + \ln(N+\frac{1}{2})+ \ln(N+\frac{1}{2})-2\ln N +o(1)\\ &= 2\ln 2 - 1 + \ln\frac{N+\frac{1}{2}}{N}+ \ln\frac{N-\frac{1}{2}}{N} +o(1)\\ &\xrightarrow[N\to\infty]{} \boxed{2\ln 2 -1} \end{align}$$ where $H_N = \sum_{n=1}^N \frac{1}{n}$ denotes the $N$-th Harmonic number, which satisfies $$ H_N = \ln N + \gamma + o(1) $$ when $N\to\infty$ ($\gamma$ being Euler's constant). We also used that by continuity of the logarithm, $\ln\frac{N\pm\frac{1}{2}}{N} \xrightarrow[N\to\infty]{} \ln 1=0$.

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    You're welcome. – 2017-01-20
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    I have one more question. When can I swap $\sum$ and $\int$ ? I see some problems that I can swap such as $\sum_{n\ge1} \frac{1}{(2n)(2n+1)}$. – 2017-01-20
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    I don't think there are full, simple answers for that, but there are some cases where this can be done: (1) power series, *inside* the radius of convergence (so here, you could have, if you had integrated from $0$ to $0.999$, for instance); (2) when you have uniform convergence of the series of function on the interval you integrate on; (3) whenever you can apply, say, the Dominated Convergence theorem to the sequence of functions defined by the partial sums of the series. (These are only three of, I don't doubt it, many cases -- also, three of the most useful, I'd say.) – 2017-01-20
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    normally I solve the problems like this by integration but I just saw the mistake of my solution. Many thanks for your advice and solution. – 2017-01-20
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    @MoNtiDeaDMoonDogs I review your proof, and it's very beautiful for me at least, but for divergence $\sum\dfrac1n$ which concludes be equal to $\ln2$ has a serious fault, so it's not applicable. good job. – 2017-01-20
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    @MyGlasses Are you talking to the OP, or to me? If the latter, I do not understand the point you are making. – 2017-01-20
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    With OP, sorry very long comments make me mistake. – 2017-01-20
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You can do this way. It is easier. Let $$ f(x)=\sum_{n=1}^\infty\frac{1}{4n^3-n}x^{2n+1}. $$ Then $f(1)=\sum_{n=1}^\infty\frac{1}{4n^3-n}$ and $$ f'(x)=\sum_{n=1}^\infty\frac{1}{(2n-1)n}x^{2n}, f''(x)=2\sum_{n=1}^\infty\frac{1}{(2n-1)}x^{2n-1}, f'''(x)=2\sum_{n=1}^\infty x^{2n-2}=\frac2{1-x^2}$$ and hence $$ f(1)=\int_0^1\int_0^s\int_0^t\frac2{1-x^2}dxdtds=2\ln2-1.$$

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    Beauty is Simplicity. (+1) – 2017-01-20