How to evaluate $\lim_{n\rightarrow \infty }n\left [ \widetilde{H_{n}}-H_{2n}+H_n \right ]$?

How to evaluate this limit with harmonic numbers

$$\lim_{n\rightarrow \infty }n\left [ \widetilde{H_{n}}-H_{2n}+H_n \right ]$$

where

$$\displaystyle \widetilde{H_{n}}=\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}$$

is the alternating harmonic number?

• Hint: Write the alternating harmonic number as a linear combination of two other harmonic numbers, and approximate all of them with natural logarithms, then use their logarithmic properties. – Lucian Jan 11 '17 at 21:21

2 Answers

One may recall the asymptotics of harmonic numbers, as $n \to \infty$, $$H_n=\log n+\gamma+\frac1{2n}+\mathcal{O}\left(\frac1{n^2} \right)$$ and one may observe that $$\widetilde{H_{n}}-\log 2=\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}-\log 2=(-1)^{n-1}\int_0^1\frac{t^n}{1+t}dt$$ Then, we may write \begin{align} &n\left [ \widetilde{H_{n}}-H_{2n}+H_n \right ] \\\\&=n\left [ \left(\widetilde{H_{n}}-\log 2\right)-\left(H_{2n}-\log (2n) \right)+\left(H_{n}-\log n \right)\right ] \\\\&=n\left [ (-1)^{n-1}\int_0^1\frac{t^n}{1+t}dt-\left(\gamma+\frac1{4n}+\mathcal{O}\left(\frac1{n^2} \right) \right)+\left(\gamma+\frac1{2n}+\mathcal{O}\left(\frac1{n^2} \right) \right)\right ] \\\\&=n (-1)^{n-1}\int_0^1\frac{t^n}{1+t}dt+\frac14+\mathcal{O}\left(\frac1{n} \right) . \end{align} Now, integrating by parts, \begin{align} n (-1)^{n-1}\int_0^1\frac{t^n}{1+t}dt&=\left. \frac{n (-1)^{n-1}t^{n+1}}{(n+1)}\frac{1}{1+t}\right|_0^1+\frac{n (-1)^{n-1}}{(n+1)}\int_0^1\frac{t^{n+1}}{(1+t)^2}\:dx\\\\ &=\frac12\frac{n}{n+1}(-1)^n+\mathcal{O}\left(\frac1{n} \right)\tag1 \end{align} and, as $n \to \infty$,

$$n\left [ \widetilde{H_{n}}-H_{2n}+H_n \right ]=\frac12\frac{n}{n+1}(-1)^n+\frac14+\mathcal{O}\left(\frac1{n} \right)$$

which does not admit a limit.

• There's an alternating $(-1)^{n-1}$ in front of the $\int_0^1 \frac{t^{n+1}}{(1+t)^2}\,dt$ too right? That should give some $\frac{(-1)^{n-1}c}{n}$ contribution too I think :o (+1) for the nice approach though :-) – r9m Apr 6 '16 at 13:11
• One may notice that $(-1)^n \times \mathcal{O}\left(\frac1{n} \right) =\mathcal{O}\left(\frac1{n} \right)$. Thanks! – Olivier Oloa Apr 6 '16 at 13:15

Ok, seems to me that there are two different problems here.

On one end : $n(-H_{2n}+H_n)$. You can rewrite this one as $-\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}$ which is, when you think of it, a way to approximate $-(\int_0^1\frac{1}{1+x} - 1)=1-\ln(2)$ (See https://en.wikipedia.org/wiki/Riemann_sum).

On the other hand, it is commonly known that, being an alternating series, $\tilde H_n$ converges towards a finite (non-zero) number, see http://pirate.shu.edu/~wachsmut/ira/numser/proofs/altharm.html for instance.

so $n\tilde H_n$ has an infinite limit.

So overall $\lim_\infty n(\tilde H_n - H_{2n} + H_n)=\infty$

EDIT : OK so seeing the other answers I just hav a slight doubt about what you mean by $H_{2n}$. Using the standard definition ($H_n=\sum_{k=1}^n\frac 1 k$) it should be $H_{2n}=\sum_{k=1}^{2n}\frac 1 k$ but it seems to me that you imply $\sum_{k=1}^{n}\frac 1 {2k}$

• The limit of $-\frac{1}{n}\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}$ is $-\int_0^1 \frac{1}{1+x}\,dx = -\ln 2$ .. – r9m Apr 6 '16 at 13:05