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The title pretty much summarizes my question. I am trying to prove the following: $$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$

I tried proving this using induction. Starting with the base case $N = 1$:

$$\displaystyle \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} = \frac{1}{1} + \frac{-1}{2} = \frac{1}{2} = \frac{1}{1+1} = \sum^{1}_{n=1} \frac{1}{N+n}.$$

My problem is the inductive step for $N+1$, starting with

$$\displaystyle \sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N+1}_{n=1} \frac{1}{(N+1)+n}.$$

And now my problem:

\begin{align} \sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} &\Leftrightarrow \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} \\[1em] &\underset{\Leftrightarrow}{\text{ind. hyp.}} \sum^{N}_{n=1} \frac{1}{N + n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} \end{align}

Is this the correct start, and if so, how do I continue?

Ali Shadhar
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phoenix
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  • The second sum ($\sum_{n = 1}^2$) ranges not from $1$ to $2$ but from $N + 1$ to $N +2$. – Antoine Jul 12 '15 at 21:33
  • I believe you want $2N$ and $N$ instead of $2(N+1)$ and $N+1$ in the start of your inductive step. – user84413 Jul 12 '15 at 21:43
  • @user84413 I don't believe that is the case--see my answer for how I "made it work." – Daniel W. Farlow Jul 13 '15 at 00:34
  • @DanielW.Farlow In your induction hypothesis, you started with $2k$ and $k$ as your upper limits of summation (instead of $2(k+1)$ and $k+1$, as the OP seemed to be starting with). – user84413 Jul 13 '15 at 14:48
  • @user84413 Ah I think I see the confusion now--you were addressing what looked like the OP starting with $2(k+1)$ and $k+1$ in the inductive hypothesis whereas I think OP was really just trying to move from the LHS of $P(k+1)$ to the RHS of $P(k+1)$ but got stuck. I think that clears it all up though. Indeed, part of the issue with inductive proofs seems to be simply writing and communicating them more than anything else. – Daniel W. Farlow Jul 13 '15 at 14:53
  • Yes, my problem was moving from the LHS of $P(k+1$ to the RHS of $P(k+1)$. To be clear, my post does not contain the induction hypothesis i was using, but the starting point of the inductive step. – phoenix Jul 13 '15 at 20:22

5 Answers5

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Claim: For all $n\geq 1$, the statement $$ P(n) : \sum_{i=1}^{2n}\frac{(-1)^{i-1}}{i}=\sum_{i=1}^n\frac{1}{n+i} $$ is true.

Note: The key to giving an inductive proof here lies in understanding how to reindex a sum; that is, the main obstacle you face in using the inductive hypothesis is realizing how to coax $\sum_{i=1}^k\frac{1}{k+i}$ out of $\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}$. To that end, first observe the following: $$ \sum_{i=1}^{k+1}\frac{1}{(k+1)+i}=\sum_{i=2}^{k+2}\frac{1}{k+i}=\sum_{i=1}^k\frac{1}{k+i}-\frac{1}{k+1}+\frac{1}{k+(k+2)}+\frac{1}{k+(k+1)}.\tag{$\dagger$} $$ Now observe that $(\dagger)$ may be rewritten as $$ \sum_{i=1}^k\frac{1}{k+i}=\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}+\frac{1}{k+1}-\frac{1}{2k+2}-\frac{1}{2k+1}.\tag{$\dagger\!\dagger$} $$ With all of that in mind, we can prove the original claim.

Proof. For the base step, we must confirm that $P(1)$ is true, something you have already done. Thus, the base case checks out.

Inductive step $P(k)\to P(k+1)$: Fix some $k\geq 1$. Assume that $$ P(k) : \color{green}{\sum_{i=1}^{2k}\frac{(-1)^{i-1}}{i}=\sum_{i=1}^{k}\frac{1}{k+i}} $$ holds. To be proved is that $$ P(k+1) : \color{blue}{\underbrace{\sum_{i=1}^{2(k+1)}\frac{(-1)^{i-1}}{i}}_{\text{LHS}}=\underbrace{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}_{\text{RHS}}} $$ follows. Beginning with the left side of $P(k+1)$, \begin{align} \text{LHS} &= \color{blue}{\sum_{i=1}^{2(k+1)}\frac{(-1)^{i-1}}{i}}\tag{definition}\\[1em] &= \color{green}{\sum_{i=1}^{2k}\frac{(-1)^{i-1}}{i}}-\frac{1}{2k+2}+\frac{1}{2k+1}\tag{by defn. of $\Sigma$}\\[1em] &= \color{green}{\sum_{i=1}^k\frac{1}{k+i}}-\frac{1}{2k+2}+\frac{1}{2k+1}\tag{by $P(k)$}\\[1em] &= \left(\color{blue}{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}+\frac{1}{k+1}-\frac{1}{2k+2}-\frac{1}{2k+1}\right)-\frac{1}{2k+2}+\frac{1}{2k+1}\tag{$\dagger\!\dagger$}\\[1em] &= \color{blue}{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}+\frac{1}{k+1}-\frac{2}{2k+2}\tag{simplify}\\[1em] &= \color{blue}{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}\tag{$k\neq -1,-1/2$}\\[1em] &= \text{RHS}\tag{definition} \end{align} one arrives at the right side of $P(k+1)$, thereby showing $P(k+1)$ is also true, completing the inductive step.

Conclusion. By mathematical induction, it is proved that for $n\geq 1$ the statement $P(1)$ is true. $\blacksquare$

Daniel W. Farlow
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1

Instead of induction, you could start with

$\displaystyle \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\sum_{n=1}^{2N}\frac{1}{n}-2\sum_{k=1}^N\frac{1}{2k}=\sum_{n=1}^{2N}\frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n}$

user84413
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you may find it easier to note that $$ S=\sum_{n=1}^{2N} (-1)^{n-1}\frac1{n} = \sum_{n=1}^{N} \left(\frac1{2n-1} - \frac1{2n} \right) $$ for then: $$ S+\sum_{n=1}^{N} \frac1{n} = \sum_{n=1}^{2N} \frac1{n} = \sum_{n=1}^{N} \left( \frac1{n}+\frac1{N+n} \right) $$

David Holden
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For LHS, use the definition of the skew harmonic number $\overline{H}_k=\sum_{n=1}^{k}\frac{(-1)^{n-1}}{n}$

$$\sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\overline{H}_{2N}=H_{2N}-H_N$$

For the RHS, $$\sum_{n=1}^N\frac{1}{N+n}=\sum_{n=1+N}^{2N}\frac{1}{n}=\sum_{n=1}^{2N}\frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n}=H_{2N}-H_N$$

Ali Shadhar
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It's easy to prove it using integration,

$$\sum_{k=1}^r\frac{1}{k+r}=\int_0^1\sum_{k=1}^r x^{k+r-1}\ dx=\int_0^1\frac{x^r-x^{2r}}{1-x}\ dx$$

$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag1$$

$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2r}(-1)^{k-1}\int_0^1 x^{k-1}\ dx=\int_0^1\sum_{k=1}^{2r}(-x)^{k-1}\ dx$$ $$=\int_0^1\frac{1-x^{2r}}{1+x}\ dx=\ln2-\int_0^1\frac{x^{2r}}{1+x}\ dx=H_{2r}-H_r\tag2$$

where the last result, follows from $\int_0^1\frac{x^{2r}}{1+x}dx=\ln2+H_r-H_{2r}$.

Hence by $(1)$ and $(2)$ , we conclude that

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

Ali Shadhar
  • 25,498