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They tell me to find the sum of the series $$\sum a_n :=\sum_{n=1}^{\infty}\frac{n+12}{n^3+5n^2+6n}$$ Since $\sum a_n$ is absolutely convergent, hence we can manipulate it the same way we would do with finite sums. I've tried splitting the general term and I get $$\frac{n+12}{n(n+2)(n+3)}=\frac{A}{n}+\frac{B}{n+2}+\frac{C}{n+3}=\frac{2}{n}+\frac{-5}{n+2}+\frac{3}{n+3}$$ and so $$\sum a_n =\sum \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}$$ Now, if I was to split the series in the sum of three different series I would get three different divergent series and so, obviously $\sum a_n$ wouldn't converge. I also suspect about being a telescopic series althought the numerators of each fraction makes it difficult to find the cancellation terms. I also know that I can rearrange the terms in my series, althought I cannot see how would this solve the problem.

If anyone could give me a hint I would really appreciate it.

HornyPigeon54
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2 Answers2

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As you said, for every $N \in \mathbb{N}^*$, one has \begin{align*}\sum_{n=1}^N a_n &=\sum_{n=1}^N \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}\\ &=2 \sum_{n=1}^N \frac{1}{n} - 5 \sum_{n=3}^{N+2} \frac{1}{n} + 3 \sum_{n=4}^{N+3} \frac{1}{n}\\ &=2 \left(1 + \frac{1}{2} + \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} \right) - 5 \left( \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} + \frac{1}{N+1} + \frac{1}{N+2} \right)\\ &\quad + 3 \left( \sum_{n=4}^N \frac{1}{n} + \frac{1}{N+1} + \frac{1}{N+2} + \frac{1}{N+3} \right)\\ &= 2 - 5 \left( \frac{1}{N+1} + \frac{1}{N+2} \right) + 3 \left( \frac{1}{N+1} + \frac{1}{N+2} + \frac{1}{N+3} \right) \end{align*}

Now just let $N$ tend to $+\infty$ to see that $$\boxed{\sum_{n=1}^{\infty} a_n = 2}$$

TheSilverDoe
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If you know how to play with harmonic numbers, it is pretty simple

$$\sum_{n=1}^p \frac 1 {n}=H_p$$ $$\sum_{n=1}^p \frac 1 {n+2}=H_{p+2}-\frac{3}{2}$$ $$\sum_{n=1}^p \frac 1 {n+3}=H_{p+3}-\frac{11}{6}$$

Now, use three times the asymptotics $$H_q=\log (q)+\gamma +\frac{1}{2 q}-\frac{1}{12 q^2}+O\left(\frac{1}{q^4}\right)$$ and continue with Taylor or long division and you will have $$\sum_{n=1}^{p}\frac{n+12}{n^3+5n^2+6n}=2-\frac{1}{p}-\frac{3}{p^2}+O\left(\frac{1}{p^3}\right)$$

Use it for $p=10$; the approximation gives exactly $1.87$ while the exact value is $\frac{1615}{858}=1.88228$ so a relative error of $0.65$%.

Claude Leibovici
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