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I attempted to evlaute the integral $I=\int_1^\infty \log(1+\frac{1}{x^2})dx$ by using the series expansion of log which gave me the following series as an answer. $\sum_{n= 0}^{\infty}\frac{(-1)^n}{(2n+1)(n+1)}$ having no real experience in evaluating such series, I compared this result with $I$ on MATLAB and concluded that they both converged to $\frac{\pi}{2}-$log$2$. Despite my lack of experience, I gave it a try anyways. Here is how my attempt went: I began by expressing the sum in two parts. $$a_n=\frac{(-1)^n}{(2n+1)(n+1)}$$so that $S=S_1+S_2$ where$$S_1=a_1+a_3+a_5+\cdot\cdot\cdot \space \space and \space S_2 =a_0+a_2++a_4+\cdot \cdot \cdot $$ which would imply $S_1=\sum_{n=0}^{\infty}\frac{1}{(4n+1)(2n+1)}$ and $S_2=-\sum_{n=0}^{\infty}\frac{1}{(4n+3)(2n+2)}$ here, I've replaced $n$ by $2n$ in the original series to get $S_1$ and $2n+1$ to get $S_2$. Then, I broke these series up into two pieces each by using partial fractions. $$S=s_1+s_2+s_3+s_4$$ where the series denoted by $s_i$ are given by the corresponding terms of the sum $$\sum_{n\ge0}\frac{1}{4n+1}-\frac{1}{2n+1}-\frac{2}{4n+3}+\frac{1}{2n+2}$$In theory, I should be able to compute all four of these sums by making use of the geometrics series. But somehow that didn't work out.My question is: What went wrong with my solution? Are there any other more practical and elegant solutions to this problem as well?

Bilge K. Aksebzeci
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2 Answers2

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We can rewrite the given summation as \begin{align*} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(n+1)} &= 2 \sum_{n=0}^\infty \left( \frac{(-1)^n}{(2n+1)} - \frac{(-1)^n}{(2n+2)} \right) \\ &= 2 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)} - \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)} \\ &= 2 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)} - \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \\ &= 2 \left( \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \cdots \right) - \left( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \cdots \right) \end{align*}

Now don't you think that we're already familiar with these 2 last infinite alternating sums? And hopefully yes! We know that $$ \arctan{x} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)} $$ And $$ \ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n} $$

Now putting $x=1$ in both of these expansions we get \begin{align*} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(n+1)} &= 2 \left( \frac{\pi}{4} \right) - \ln(2) \\ &= \frac{\pi}{2} - \ln(2) \end{align*}

Lucky Chouhan
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    @GEdgar The series are convergent due to the alternating series test (though it should be mentioned before splitting up, I agree), and the values for the desired sums are correct, however the issue I see is that you can't just "put $x = 1$ in the expansions" since the convergence radii are $1$, so Chouhan would need to justify that using other means, or to just assume the values as known. – Bruno B Jul 29 '23 at 09:50
  • Also to justify re-arranging the conditionally convergent series like this. $$2 \sum_{n=0}^\infty \left( \frac{(-1)^n}{(2n+1)} - \frac{(-1)^n}{(2n+2)} \right) \ = 2 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)} - \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)}$$ It is OK since the first series is actually absolutely convergent? – GEdgar Jul 29 '23 at 09:51
  • @GEdgar There's no re-arranging here, it's just a linear combination of series, so you only need both series to converge to split them up, but they don't have to converge absolutely. Absolute convergence would be needed if there were a rearrangement as in you change the order of summation by moving around an infinite amount of terms of one series, i.e. for stuff like $\sum u_n = \sum u_{\varphi(n)}$ with $\varphi : \mathbb{N} \to \mathbb{N}$ a bijection. – Bruno B Jul 29 '23 at 11:35
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    Or if you want to sum $\sum_n u_n$ as $\sum_n \sum_{k = p_{n-1} + 1}^{p_n} u_k$ with $(p_n)_n$ possibly unbounded, then absolute convergence is also needed. – Bruno B Jul 29 '23 at 11:43
  • @BrunoB yeah! I just distributed the summation. And so sorry I put $x=1$ because I have seen in many places doing like this as in Leibniz's formula for $\pi/4$ we just need to put $x=1$ in the expansion of $\tan^{-1}(x)$ and in 2nd one it is well know that alternating harmonic series converges to $\ln(2)$ so that's how I got both sums. – Lucky Chouhan Jul 29 '23 at 11:43
  • @BrunoB But thank you for pointing it out. I should have done it more rigorously. I'm going to learn more about radius of convergence. – Lucky Chouhan Jul 29 '23 at 11:44
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    You can justify the values you get here thanks to Abel's theorem for power series and the alternating series test telling you the series converge, fortunately. But you can't always apply it, and there's not always continuity at a point of the boundary so you have to be careful. – Bruno B Jul 29 '23 at 11:52
  • @BrunoB Yeah sure! after reading that article I'm going to update my answer. Again thank you for your constructive criticism <3 – Lucky Chouhan Jul 29 '23 at 11:56
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It would work if you consider the partial sums first : writing, as you did $$S_p=\sum_{n= 0}^{p}\frac{(-1)^n}{(2n+1)(n+1)}=S_{1p}-S_{2p}$$ $$S_{1p}=\sum_{n=0}^{p}\frac{1}{(4n+1)(2n+1)}=2\sum_{n=0}^{p}\frac{1}{4 n+1}-\sum_{n=0}^{p}\frac{1}{2 n+1}$$ $$S_{2p}=\sum_{n=0}^{p}\frac{1}{(4n+3)(2n+2)}=2\sum_{n=0}^{p}\frac{1}{4 n+3}-\sum_{n=0}^{p}\frac{1}{2 n+2}$$ Now, using four times $$\sum_{n=0}^{p}\frac{1}{a n+b}=\frac 1a \left(\psi \left(p+1+\frac{b}{a}\right)-\psi\left(\frac{b}{a}\right)\right)$$ you will have $S_p$.

Now, using the asymptotics $$\psi(q+c)=\log (q)+\frac{2 c-1}{2 q}-\frac{6 c^2-6 c+1}{12 q^2}+O\left(\frac{1}{q^3}\right)$$ you will have more than the limit itself.

Claude Leibovici
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