To Prove $\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k}\right).$

Question

Prove That $\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k}\right).$

My Approach I was solving:

$$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}= \left(1+\frac 1 3 + \frac 1 5 +\ldots \right)-\left(\frac 1 2 + \frac 1 4 +\ldots \right) =\sum_{n=1}^{\infty}\frac{1}{2n-1}- \sum_{n=1}^{\infty}\frac{1}{2n}$$ Searching On Mathstack i found This

I know that $ \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}$ = $\ln2$ But i don't know how to prove it using this method.

Answer 1

Both series$$\sum_{n=1}^\infty\frac1{2n-1}\text{ and }\sum_{n=1}^\infty\frac1{2n}$$diverge. Therefore, the equality$$\sum_{n=1}^\infty\frac1{2n-1}-\sum_{n=1}^\infty\frac1{2n}=\ln2$$makes no sense.

---

Note: What I wrote was an answer to the problem of proving that$$\sum_{n=1}^\infty\frac1{2n-1}-\sum_{n=1}^\infty\frac1{2n}=\ln2.$$The question was edited afterwards.

Answer 2

Perhaps you're confusing this with,

$$\sum_{n=1}^\infty \left(\frac 1 {2n-1} - \frac 1 {2n}\right)$$

Note that both $\sum \frac 1 {2n-1}$ and $\sum \frac 1 {2n}$ are divergent from the divergence of the harmonic series, so to "split up" the summand does not make mathematical sense.

However, to say,

$$\lim_{k \to \infty} \left(\sum_{n=1}^k \frac 1 {2n-1} - \sum_{n=1}^k \frac 1 {2n}\right)$$

Is fine. It is however not equivalent to what you write. Note that just as we can't split the above sum using additivity as both sums diverge, we cannot split the above limit using additivity as both terms increase without bound.

Answer 3

As said in the other answer, it makes no sense to split the convergent sum into two divergent sums.

However, we can look at what happens for partial sums.

Notice that $\quad -\dfrac 1{2k}=\dfrac 1{2k}-\dfrac 1k\quad$ so

$\begin{align}\sum\limits_{n=1}^{2N}\dfrac{(-1)^{n-1}}n &=\sum\limits_{n=1}^N\left(\dfrac 1{2n-1}-\dfrac 1{2n}\right) =\sum\limits_{n=1}^N\left(\dfrac 1{2n}+\dfrac 1{2n-1}-\dfrac 1{n}\right)\\\\ &=\sum\limits_{n=1}^{2N} \dfrac 1n-\sum\limits_{n=1}^N\dfrac 1n =\sum\limits_{n=N+1}^{2N}\dfrac 1n=\sum\limits_{n=1}^{N}\dfrac 1{n+N}\\\\ &=\dfrac 1N\sum\limits_{n=1}^{N}\dfrac 1{1+\frac nN}\xrightarrow{\text{Riemann sum}}\int_0^1\dfrac{dt}{1+t}=\ln(2)\end{align}$

Answer 4

It may all depend on the meaning you give to things...and on the way you write things that are agreed by almost everybody else.

If we take a finite sum of real numbers then we can use commutativity, associativity and etc. with, so we could agree that

$$\sum_{k=1}^n\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}=\left(1+\frac13+\ldots+\frac1{2n-1}\right)-\left(\frac12+\frac14+\ldots+\frac1{2n}\right)=$$

$$=1-\frac12+\frac13-\frac14+\ldots+\frac1{2n-1}-\frac1{2n} (**)$$

and then indeed we get

$$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}\right)=\lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\ldots+\frac1{2n-1}-\frac1{2n}\right)=\log2$$

since we have the power series

$$\log(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}n$$

which is valid for $\;|x|\le1\;$ (why?), and thus with $\;x=1\;$ we get the above result.

Observe though that we could have agreed in $\;(**)\;$ that, for example:

$$\left(1+\frac13+\ldots+\frac1{2n-1}\right)-\left(\frac12+\frac14+\ldots+\frac1{2n}\right)=$$

$$=1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\ldots=\frac12\log2$$ and we get something different! This can happen, and actually does happen, when we rearrange the terms of a conditionally convergent series, as the alteranting harmonic series is, which you can rearrange it to get any sum you want, up to and including $\;\pm\infty\;$ !

You may want to read about "Riemann Theorem" in series. Google it...it is fascinating.