# Convergence and limit of $a_1 - a_2 +a_3-a_4 + \cdots$, where $a_n=\frac{1+\frac{1}{2}+\frac{1}{3} + \cdots +\frac{1}{n}}{n}$ [closed]

For $$n=1,2,3,...$$, define $$a_n=\frac{1+\frac{1}{2}+\frac{1}{3} + \cdots +\frac{1}{n}}{n}$$ Consider the series $$S= a_1 - a_2 +a_3-a_4 + \cdots$$. Prove that the series converges conditionally and what is its limit?

Using Liebnitz test I have proved the convergence of S. Using Comparison test I have proved that S is not absolutely convergent. How do I find the sum of the series S? Thank you for your help.

• +1 to your question. However, some mathSE reviewers prefer that the posted question shows some analysis. Therefore, please edit your question to include (at least some) explicit details of how you reached your intermediate conclusions. – user2661923 May 23 at 13:49
• Personally, if I was attacking this problem, my first step would be (perhaps on scratch paper) to explicitly construct $a_1, a_2, \cdots, a_6$ and then construct $a_1 - a_2 + a_3 - a_4 + a_5 - a_6$. I would then look for a pattern, and try to form a hypothesis that represents generalizing this pattern. Then, I would try to prove the hypothesis, perhaps via induction. – user2661923 May 23 at 13:53
• @user2661923 $a_n is monotonic decreasing (by induction) and converging to zero since$\frac{1}{n}$converges to zero. To prove that it doesn't converge absolutely compare with$sum(1/n)$. The limit is infinite and hence S diverges. – Lawrence Mano May 23 at 15:58 • I am surprised by your comment. My 2nd comment focused on assuming that the sequence was conditionally convergent, which you indicated that you could prove. The idea behind my 2nd comment was to identify the exact value that the alternating series converged to by exploring$a_1 - a_2 + a_3 - a_4 + a_5 - a_6.– user2661923 May 23 at 16:30 • @LawrenceMano You claim you tried two things. Can we see your work, so evaluate, in case you made a mistake? – amWhy Jun 12 at 22:33 ## 2 Answers The series of interest is given by $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}H_n$$ where $$H_n =\sum_{k=1}^n \frac1k$$. Inasmuch as $$H_n=\log(n)+\gamma +O(1/n)$$, the series converges. Therefore we can write \begin{align} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}H_n&=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sum_{k=1}^n \frac1k\\\\ &=\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty \frac{(-1)^{n+1}}{n}\\\\ &=\sum_{k=1}^\infty \sum_{n=0}^\infty \frac{(-1)^{n+k+1}}{k(n+k)}\\\\ &=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}-\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{k(n+k)}\\\\ &=\frac{\pi^2}{12}-\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{k(n+k)}\tag1\\\\ &=\frac{\pi^2}{12}-\sum_{k=1}^\infty \sum_{n=1}^\infty \left(\frac{(-1)^{n+k}}{nk}-\frac{(-1)^{n+k}}{n(n+k)}\right)\tag2 \\\\ &=\frac{\pi^2}{12}-\log^2(2)+\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{n(n+k)}\tag3 \end{align} The interchange of the order of summation is straightforward to justify and is left to the reader to verify. In going from $$(1)$$ to $$(2)$$ we used partial fraction expansion. Now, not that the series on the right-hand side of $$(3)$$ is identical to the series on the right-hand side of $$(1)$$ (indices interchanged). Hence, we see that $$2\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{k(n+k)}=\log^2(2)$$ from which we find that $$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}H_n=\frac{\pi^2}{12}-\frac12\log^2(2)}$$ And we are done. • In the last equation, you probably mean\frac{(-1)^{n+1}}n$– Kenta S May 23 at 14:59 • @KentaS Indeed. Thank you for the catch. I've edited accordingly. – Mark Viola May 23 at 15:16 • Can you please explain equation (3). – Lawrence Mano May 23 at 16:12 • Sure. The first double series is the square of the series$\sum_{n=1}^\infty \frac{(-1)^n}{n}$. This singe series is$-\log(2)\$. Now square this. – Mark Viola May 23 at 16:15

First of all, the generating function of the harmonic numbers is: $$\sum_{n\ge1}H_nz^{n-1}=-\frac{\ln(1-z)}{z(1-z)}=-\frac{\ln(1-z)}{1-z}-\frac{\ln(1-z)}z,$$ absolutely convergent for $$|z|<1$$. Integrating, we obtain: \begin{align} \sum_{n\ge 1}\frac1{n}H_nz^n&=-\int_0^z\left[\frac{\ln(1-t)}{1-t}+\frac{\ln(1-t)}{t}\right]dt\\ &=\frac12(\ln(1-z))^2+\int_0^z\frac{\ln(1-t)}tdt\ (|z|<1).\end{align} Now, taking the limit $$z\to-1$$, we obtain: \begin{align}\sum_{n\ge 1}\frac{(-1)^n}nH_n&=\frac12(\ln2)^2+\int_0^{-1}\frac{\ln(1-t)}tdt\\ &=\frac12(\ln2)^2-\frac{\pi^2}{12},\end{align}

where the last integral can be evaluated with the same method as done here.

P.S. The exchange of the limit and integral can be justified as follows.

We have

\begin{align}\sum_{n\ge 1}\frac1n H_n(-1+\varepsilon)^n&=\sum_{k\ge0}\left(\frac1{2k+1}H_{2k+1}(-1+\epsilon)^{2k+1}+\frac1{2k+2}H_{2k+2}(-1+\epsilon)^{2k+2}\right)\\ &=\sum_{k\ge0}\left(\frac1{2k+1}H_{2k+1}-\frac1{2k+2}H_{2k+2}+\epsilon\frac1{2k+2}H_{2k+2}\right)(-1+\epsilon)^{2k+1}.\\ \end{align} Now, for small enough $$\epsilon>0$$: \begin{align}\left|\left(\frac1{2k+1}H_{2k+1}-\frac1{2k+2}H_{2k+2}+\epsilon\frac1{2k+2}H_{2k+2}\right)(-1+\epsilon)^{2k+1}\right|&\le\left(\frac1{2k+1}H_{2k+1}-\frac1{2k+2}H_{2k+2}\right)+\epsilon(1-\epsilon)^{2k+1}\frac1{2k+2}H_{2k+2}\\&\le\left(\frac1{2k+1}H_{2k+1}-\frac1{2k+2}H_{2k+2}\right)+\frac1{(2k+2)^2}H_{2k+2}.\end{align} Now, denoting the RHS as $$a_k$$, the sum $$\sum a_k$$ converges, so by the dominated convergence theorem we can exchange the limit and the sum.

• How does one justify interchanging the limit with the series? – Mark Viola May 23 at 16:09