Show that for all integers greater or equal to $1$, the alternating harmonic series can be written as $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{1}{k+1}$

Question

The original question:

Show that $\forall \, n\in \mathbb{Z}, n\geq 1$ the equation below holds

$$\begin{align} 1-\frac12 + \frac13 - \frac14 + \cdots + \frac{1}{2n-1} - \frac{1}{2n} =& \frac{1}{n+1} + \frac{1}{n+2} \\ & +\cdots + \frac{1}{2n-1} + \frac{1}{2n} \end{align}$$

In case further details are needed, here's the screenshot of the whole question. I took the question from How to Integrate It regarding Darboux Sum.

When I saw the word integers, my mind was sent to Mathematical Induction. However, I noticed that they are series and I doubt it would be helpful since if I plug in $n=1$ as an initiation of mathematical induction, it would fail at the term $\dfrac{1}{2n-2}$. So perhaps the writer of the book wants me to prove with something else? How?

I also noticed that the term on the left is an alternating Harmonic Series which originally I believe I could rewrite them as follows:

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

I was in doubt whether include the limit sign as $n$ approaches to infinity or not since from the question (c) the author told me so. Any help would be appreciated. Thank you.

This question is similar to: Proof by induction: $\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. When plugging in $n=1$, it does work since there's no $\frac{1}{2n-1}$ term, but instead just $\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{1}-\frac{1}{2}$. Also, note the ... — John Omielan, Nov 18 '24 at 05:41
(cont.) duplicate was found using an Approach0 search. It also listed quite a few other duplicates, and near duplicates, such as Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$, ... — John Omielan, Nov 18 '24 at 05:45
(cont.) Prove Catalan's Identity, Prove that for all $n \ge 1$, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{1}{2n}$, Help with following equality, Proof that $\sum_{i=1}^n \frac{1}{n+i}$ is bounded, Convergence of $\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$, etc. — John Omielan, Nov 18 '24 at 05:48

score 1 · Accepted Answer · answered Nov 18 '24 at 05:49

You can use induction. The base case, for $n=1$ is $1-\frac 12=\frac 12$, which is true. Now assume it is true up to $k$. For $n=k+1$ we have $$1-\frac 12 + \frac 13 - \frac 14 \ldots +\frac 1{2k+1}-\frac 1{2k+2}=\\ \frac 1{k+1}+\frac 1{k+2}+\ldots \frac 1{2k}+\frac 1{2k+1}-\frac 1{2k+2}\\ =\frac 1{k+2}+\frac 1{k+3}+\ldots \frac 1{2k+1}+\frac 1{2k+2}$$ where in the first $=$ we plugged in the induction assumption and in the second we used $\frac 1{k+1}-\frac 1{2k+2}=\frac 1{2k+2}$

Show that for all integers greater or equal to $1$, the alternating harmonic series can be written as $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{1}{k+1}$

1 Answers1