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

Question

How to prove that

$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}\tag1$$

We know that both sides are equal to $H_{2r}-H_r$ but I am trying to convert the left side to the right side using only series manipulations and without going through $H_{2r}-H_r$

There is nothing I could try but we know that $$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1+r}^{2r}\frac{1}{k}$$

What next? Thank you.

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By the way, its easy to prove it using integration,

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

$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag2$$

and we proved here

$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=H_{2r}-H_r\tag3$$

Hence by $(2)$ and $(3)$ , $(1)$ is proved

Answer 1

Here's another direct proof.

The difference between r.h.s. and l.h.s is

$$\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}-\sum_{k=1}^r\frac{1}{k+r} \\=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}-\sum_{k=r+1}^{2r}\frac{1}{k} \\=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}-\sum_{k=1}^{2r}\frac{1}{k}+\sum_{k=1}^{r}\frac{1}{k} \\=\sum_{k=1}^{2r}\frac{1}{k}\left((-1)^{k-1}-1\right)+\sum_{k=1}^{r}\frac{1}{k} \\\overset{k\to 2m}=\sum_{m=1}^{r}\frac{1}{2m}\left(-2\right)+\sum_{k=1}^{r}\frac{1}{k}=0$$

Hence l.h.s = r.h.s. QED

Answer 2

Let

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

To show that $u(r) = v(r)$ it is enough to show that $u(1) = v(1)$ and $u(r+1)-u(r) =v(r+1)-v(r) $.

$u(1) =\frac12$ and $v(1) =1-\frac12 =\frac12$.

$v(r+1)-v(r) =\sum_{k=2r+1}^{2r+2}\frac{(-1)^{k-1}}{k} =\frac1{2r+1}-\frac1{2r+2} =\frac1{(2r+1)(2r+2)} $.

$\begin{array}\\ u(r+1)-u(r) &=\sum_{k=1}^{r+1}\frac{1}{k+r+1}-\sum_{k=1}^r\frac{1}{k+r}\\ &=\sum_{k=r+2}^{2r+2}\frac{1}{k}-\sum_{k=r+1}^{2r}\frac{1}{k}\\ &=\sum_{k=r+2}^{2r}\frac{1}{k}+\frac{1}{2r+1}+\frac{1}{2r+2}-(\frac{1}{r+1}+\sum_{k=r+2}^{2r}\frac{1}{k})\\ &=\frac{1}{2r+1}+\frac{1}{2r+2}-\frac{1}{r+1}\\ &=\frac{1}{2r+1}-\frac{1}{2r+2}\\ &=\frac{1}{(2r+1)(2r+2)}\\ &=v(r+1)-v(r)\\ \end{array} $

Answer 3

As has been pointed out by @Greg Martin, one way to get the result you desire using only manipulations of series would be to essentially run the proof of the result I gave here in reverse.

Starting with $\sum_{k = 1}^r \frac{1}{k + r}$ reindexing $k \mapsto k - r$ we have \begin{align} \sum_{k = 1}^r \frac{1}{k + r} &= \sum_{k = r + 1}^{2r} \frac{1}{k}\\ &= \sum_{k = 1}^{2r} \frac{1}{k} - \sum_{k = 1}^n \frac{1}{k}\\ &= \left (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{r} \right )\\ &= 1 + \left (\frac{1}{2} - 1 \right ) + \frac{1}{3} + \left (\frac{1}{4} - \frac{1}{2} \right ) + \frac{1}{5} + \left (\frac{1}{6} - \frac{1}{3} \right ) + \cdots\\ &\qquad \cdots + \frac{1}{2n - 1} + \left (\frac{1}{2n} - \frac{1}{n} \right )\\ &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}\\ &= \sum_{k = 1}^{2r} \frac{(-1)^{k + 1}}{k}, \end{align} as desired.

Of course this is just $H_{2r}$ and $H_r$ in disguise. So let us try again starting from the other side. \begin{align} \sum_{k = 1}^{2r} \frac{(-1)^{k + 1}}{k} &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2r - 1} - \frac{1}{2r}\\ &= \left (1 - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \cdots + \left (\frac{1}{2r - 1} - \frac{1}{2r} \right )\\ &= \left (1 + \frac{1}{2} - 2 \cdot \frac{1}{2} \right ) + \left (\frac{1}{2} + \frac{1}{4} - 2 \cdot \frac{1}{4} \right ) + \cdots\\ & \qquad \cdots + \left (\frac{1}{2r - 1} + \frac{1}{2r} - 2 \cdot \frac{1}{2r} \right )\\ &= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - 2 \left (\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2r} \right )\\ &= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - \left (1 + \frac{1}{2} + \cdots + \frac{1}{r} \right )\\ &= \frac{1}{r + 1} + \frac{1}{r + 2} + \cdots + \frac{1}{2r}\\ &= \sum_{k = 1}^r \frac{1}{k + r}, \end{align} which is essentially the same thing.

Finally, this identity is known as the Botez-Catalan identity.