1

Prove:

$$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$$

where $H_n=\sum\limits_{k=1}^n\frac1k$

For the $x^k$ i tried $x^k=((x-1)+1)^k$ and decompose it into binomial expansion but I got dual sum at the one side one sum at the other side

For the second method:

$$\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}-\sum\limits_{k=1}^n \frac{x^k}{k}=H_n$$

and try to show

$$\binom{n}{k}(x-1)^k-x^k=1$$

Either the problem is wrong or I cannot do it.

Jale'de jaled
  • 780

2 Answers2

1

I think the identity does not hold. Using $$ x^k=[1+(x-1)]^k=\sum_{i=0}^k\binom{k}{i}(1-x)^i=1+\sum_{i=1}^k\binom{k}{i}(1-x)^i $$ one has \begin{eqnarray} &&\sum\limits_{k=1}^n \frac{x^k}{k}\\ &=&\sum\limits_{k=1}^n\frac{1}{k}\bigg(1+\sum_{i=1}^k\binom{k}{i}(1-x)^i\bigg)\\ &=&H_n+\sum\limits_{k=1}^n\sum_{i=1}^k \binom{k}{i} \frac{(x-1)^i}{k}\\ &\neq&H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}. \end{eqnarray}

xpaul
  • 44,000
  • If in your second-to-last line you switch the order of summation, you get $$\sum_{k = 1}^{n} \frac{x^k}{k} = H_n + \sum_{i = 1}^{n} \frac{(x-1)^{i}}{i} \sum_{k = i}^{n} \binom{k-1}{i-1},.$$ – Daniel Fischer Nov 11 '20 at 16:36
  • Yes, you are right. I just wanted to keep the same form. – xpaul Nov 11 '20 at 16:39
  • Sorry, I was too opaque. Use the so-called Hockey-stick identity, and then rename $i$ to $k$. – Daniel Fischer Nov 11 '20 at 16:54
-2

This identity does hold true

express both the RHS and LHS as integrals

then interchange the integrals and summations.

Hope this helps

user848414
  • 1