3

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.

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

Ali Shadhar
  • 25,498
  • 2
    Your linked question gives a series-manipulation proof that the right-hand side is $H_{2r}-H_r$, and the left-hand side is simply $$\sum_{k=r+1}^{2r} \frac1k = \sum_{k=1}^{2r} \frac1k - \sum_{k=1}^{r} \frac1k = H_{2r}-H_r$$ by definition. You can just combine the two computations to get the proof you want. – Greg Martin Feb 17 '20 at 21:05
  • @Greg Martin I think you misunderstood or my wording was confusing. I want to convert the left side to the right side using series manipulation without going through $H_{2r}-H_r$ – Ali Shadhar Feb 17 '20 at 21:14
  • Yes. The linked answer has the computation$$\text{\rm (RHS of (1))}{} = A = B=C=D=E=H_{2r}-H_r,$$and in my previous comment is the computation $$\text{\rm (LHS of (1))} = F = H_{2r}-H_r.$$ Therefore the proof you are looking for is $$\text{\rm(LHS of (1))} = F = E=D=C=B=A=\text{\rm(RHS of (1))}.$$ (Indeed $E$ and $F$ are the same expression, but that's not relevant.) – Greg Martin Feb 17 '20 at 21:16
  • @Greg Martin I edited my question. I think its less confusing now. – Ali Shadhar Feb 17 '20 at 21:20
  • I think induction will work. – mjw Feb 17 '20 at 21:26
  • @GregMartin let's see what the others say. Thanks though. – Ali Shadhar Feb 17 '20 at 21:31
  • @ Ali Shather Nice problem (I would do it inductively). Just a remark: it is interesting that the equation does not hold for the analytic continuations of both sides to real $r$ unless $r$ is integer. Indeed, the closed form of the lhs is $H_{2r}-H_{r}$, and that of the lhs is $(-1)^{2 r+1} \Phi (-1,1,2 r+1)+\log (2)$. Now due to the facor $(-1)^{2r+1}$ the rhs becomes complex for non integer $r$ so that the equality is no longer valid. – Dr. Wolfgang Hintze Feb 18 '20 at 09:45

3 Answers3

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} $

marty cohen
  • 107,799
1

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.

omegadot
  • 11,736
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

Dr. Wolfgang Hintze
  • 12,465
  • Thank you Dr. but I said in the question that I need to manipulate the LHS to get the RHS. To be more clear I want to see manipulations like Cauchy product, exchanging double series , manipulating the series boundaries and other tricks. – Ali Shadhar Feb 19 '20 at 00:16
  • @ Ali Shather Sorry, but a closer look at my exposition will show you that I mainly manipulated your LHS, as requested in the OP. – Dr. Wolfgang Hintze Feb 19 '20 at 06:40
  • Sorry Wolf just saw your reply... actually your solution is the one I'm looking for. Perfect +1 – Ali Shadhar May 24 '20 at 20:05