Find $\lim_{n\to \infty} (1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})$

Question

Find : $\lim_{n\to \infty} (1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})$ .

My Approach:

Let $x_n=(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})$ .

I know that $\gamma_{n}=\sum_{n=1}^n \frac{1}{n} -\log(n)$ . And $<\gamma_n>$ converges to Euler's constant i.e. $\gamma$ which lies between $0.3$ and $1$ .

Using this fact I found ,

$x_n=1+\gamma_{2n-1}+\log(2n-1)-\gamma_{2n}-\log(2n)$

since $\gamma_{2n-1}$ and $\gamma_{2n}$ are subsequences of convergent sequence $<\gamma_n>$ , so they converges to same limit as $n$ goes to $\infty$ .

Hence , $\lim_{n\to \infty} x_n= 1$ .

Question:

To check whether I'm wrong or right . Actually I don't have answers , Please help!

EDIT: $(1)$ My approach is wrong.

Answer 1

To answer the question of what you did wrong in your approach specifically, it's just that your arithmetic is wrong. You write (essentially) $x_n=1+\gamma_{2n-1}+\log(2n-1)-\left(\gamma_{2n}+\log(2n)\right)$; your idea of writing $x_n$ as a difference of harmonic terms is broadly correct, but let's look at what's going on here. I'll write (using your notation) $H_n=\gamma_n+\log(n)=\sum_{i=1}^n\frac1i$ for the harmonic numbers; then what you've written is $1+H_{2n-1}-H_{2n}$. But almost all of the terms cancel out of this sum, because it's not the case that $H_{2n-1}=1+\frac13+\frac15+\ldots+\frac1{2n-1}$ and $H_{2n}=\frac12+\frac14+\ldots+\frac1{2n}$; instead, $H_{2n-1}=1+\frac12+\frac13+\ldots$ and similarly for $H_{2n}$, so what you've written is just $1-\frac1{2n}$.

Instead, to use the approach you're trying, you need to start with $1+\frac12+\frac13+\ldots+\frac1{2n}$ and then subtract the even terms twice :once to 'eliminate' them from the harmonic sum, yielding $1+\frac13+\frac15+\ldots$, and then a second time to 'add' the negative terms and give the sum $1-\frac12+\frac13-\frac14+\ldots$ that you want. This means that your sum is $$\begin{eqnarray} 1+\frac12+\frac13+\ldots+\frac1{2n}-2(\frac12+\frac14+\ldots+\frac1{2n}) \\ =1+\frac12+\ldots+\frac1{2n}-(\frac22+\frac24+\frac26+\ldots+\frac2{2n}) \\ =1+\frac12+\ldots+\frac1{2n}-(1+\frac12+\frac13+\ldots+\frac1n) \\ =H_{2n}-H_n \end{eqnarray}$$. Can you finish from here?

Answer 2

Hint: The Taylor expansion for $\log (x+1)$ around 0 is $$x-x^2/2+x^3/3-\cdots = \sum_{n\ge1}\frac{(-1)^{n+1}x^n}{n}$$

Answer 3

Your idea is actually correct, you made a mistake in the computations:

Note that $$x_n=1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n}= \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}\right)-2 \left( \frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n}\right)\\=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}\right)- \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\\=\gamma_{2n} +\ln(2n)-\gamma_n-\ln(x)=\gamma_{2n}-\gamma_n+\ln(2)$$

Alternate solution The following is a classical alternate sulution:

Prove by induction the (well known identity):

$$1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$$ and identify the RHS as a Riemann sum.