What is the sum this series involving sum of reciprocals of natural numbers?

Question

Let $S =^nC_1 - (1+\frac{1}{2}) ^nC_2 + (1+\frac{1}{2}+\frac{1}{3}) ^nC_3 + … (-1)^{n-1}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{n}) (^nC_n)$ I have to prove that $S = \frac{1}{n}$, I thought of writing the general term like this,

$T_r=(-1)^{r-1}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{r}) (^nC_r)$ and then multiplying each term by $^nC_r$ and then take sum of each, but that led me nowhere. I tried the reversal method too, but the terms weren't adding up. I want to know if there is a easy way of doing this. Thanks.

Answer 1

$$S_n=\sum_{k=0}^n (-1)^k H_k {n \choose k}, H_k=1+1/2+1/3+....+1/k$$ $$H_k=\int_{0}^1 \frac{t^{k}-1}{t-1} dt$$ Then $$S_n=\int_{0}^{1} \frac{dt}{t-1}\sum_{k=0}^n(-1)^k {n \choose k}(t^{k}-1)$$ $$S_n=\int_{0}^{1} \frac{dt}{t-1}\sum_{k=0}^n \left((-1)^k {n \choose k}(t^{k}-1)\right)$$ $$\int_{0}^{1} \frac{dt}{(t-1)} [(1-t)^{n}-0]=-\int_{0}^{1}(1-t)^{n-1}dt=\frac{1}{n}$$

Answer 2

You get it by multiplying $ (1-x)^n(1+x+x^2........)$

See the video of the solution for the problem (https://youtu.be/F-OmxeQxSFs)

Answer 3

$$S_n=\sum_{i=1}^n\left((-1)^{i-1}\left(\sum_{k=1}^i\frac{1}{k}\right)\binom{n}{i}\right)$$

So, $$S_n-S_{n-1}=\sum_{i=1}^n\left((-1)^{i-1}\left(\sum_{k=1}^i\frac{1}{k}\right)\binom{n}{i}\right)-\sum_{i=1}^{n-1}\left((-1)^{i-1}\left(\sum_{k=1}^i\frac{1}{k}\right)\binom{n-1}{i}\right)$$$$=\sum_{i=1}^{n-1}\left((-1)^{i-1}\left(\sum_{k=1}^i\frac{1}{k}\right)\left(\binom{n}{i}-\binom{n-1}{i}\right)\right)+(-1)^{n-1}\left(\sum_{j=1}^n\frac{1}{j}\right)$$

Then, by this recursion relation, we get:

$$S_n-S_{n-1}=\sum_{i=1}^{n-1}\left((-1)^{i-1}\left(\sum_{k=1}^i\frac{1}{k}\right)\binom{n-1}{i-1}\right)+(-1)^{n-1}\left(\sum_{j=1}^n\frac{1}{j}\right)$$$$=\sum_{i=1}^{n-1}\left((-1)^{i-1}\left(\sum_{k=1}^{i-1}\frac{1}{k}\right)\binom{n-1}{i-1}\right)+\sum_{i=1}^{n-1}\left((-1)^{i-1}\frac{1}{i}\binom{n-1}{i-1}\right)+(-1)^{n-1}\left(\sum_{j=1}^n\frac{1}{j}\right)$$$$=-\sum_{i=2}^{n-1}\left((-1)^{i-2}\left(\sum_{k=1}^{i-1}\frac{1}{k}\right)\binom{n-1}{i-1}\right)+\sum_{m=1}^{n-1}\left((-1)^{m-1}\frac{1}{m}\binom{n-1}{m-1}\right)+(-1)^{n-1}\left(\sum_{j=1}^n\frac{1}{j}\right)$$

If you substitute $l$ for $i-1$:

$$S_n-S_{n-1}=-\sum_{l=1}^{n-2}\left((-1)^{l-1}\left(\sum_{k=1}^l\frac{1}{k}\right)\binom{n-1}{l}\right)+\sum_{i=1}^{n-1}\left((-1)^{i-1}\frac{1}{i}\binom{n-1}{i-1}\right)+(-1)^{n-1}\left(\sum_{j=1}^n\frac{1}{j}\right)$$$$=-\sum_{l=1}^{n-1}\left((-1)^{l-1}\left(\sum_{k=1}^l\frac{1}{k}\right)\binom{n-1}{l}\right)+(-1)^{n-2}\left(\sum_{k=1}^{n-1}\frac{1}{k}\right)+\sum_{i=1}^{n-1}\left((-1)^{i-1}\frac{1}{i}\binom{n-1}{i-1}\right)+(-1)^{n-1}\left(\sum_{j=1}^n\frac{1}{j}\right)$$$$=\sum_{l=1}^{n-1}\left((-1)^{l-1}\left(\sum_{k=1}^l\frac{1}{k}\right)\binom{n-1}{l}\right)+\sum_{i=1}^{n-1}\left((-1)^{i-1}\frac{1}{i}\binom{n-1}{i-1}\right)+\frac{(-1)^{n-1}}{n}$$$$=-S_{n-1}+\sum_{i=1}^{n-1}\left((-1)^{i-1}\frac{1}{i}\binom{n-1}{i-1}\right)+\frac{(-1)^{n-1}}{n}$$

Cancelling $-S_{n-1}$:

$$S_n=\sum_{i=1}^{n-1}\left((-1)^{i-1}\frac{1}{i}\binom{n-1}{i-1}\right)+\frac{(-1)^{n-1}}{n}$$

Then, given that $\binom{n-1}{i-1}=\frac{(n-1)!}{(n-i)!(i-1)!}$, $\binom{n}{i}=\frac{n!}{(n-i)!i!}=\frac{n}{i}\binom{n-1}{i-1}$, so:

$$S_n=\frac{1}{n}\sum_{i=1}^{n-1}\left((-1)^{i-1}\binom{n}{i}\right)+\frac{(-1)^{n-1}}{n}$$$$=-\frac{1}{n}\left(\sum_{i=0}^n\left(1^{n-i}(-1)^i\binom{n}{i}\right)\right)+\frac{1+(-1)^n}{n}+\frac{(-1)^{n-1}}{n}$$

By the binomial theorem, then: $$S_n=\frac{-(1+(-1))^{n-1}}{n}+\frac{1+(-1)^n}{n}+\frac{(-1)^{n-1}}{n}$$$$S_n=\frac{1}{n}$$

(please comment or edit for any corrections)