Finite Binomial Sum: Proving 1 + 1/2 + 1/3 + ... + 1/n

[DreamWeaver]

Show that

$$\sum_{j=1}^{j=n}\binom{n}{j} \frac{(-1)^{j+1}}{j} = 1 +\frac{1}{2} +\frac{1}{3} + \cdots +\frac{1}{n}$$

 

[lfdahl]

A small hint

Spoiler

- is missing

 

[DreamWeaver]

A small hint

Spoiler

- is missing

Really sorry, Ifdahl!
Keep thinking I'll have time, and then I'm unexpectedly busy... Bad mammal!

Sorry...

 

[alyafey22]

Nice question. I think I know how to approach it. I'll write the solution later.

 

[alyafey22]

Spoiler

$$\sum_{j=0}^n {n \choose j}(-x)^j=(1-x)^n
$$

$$\sum_{j=1}^n {n \choose j}(-1)^{j}x^{j-1}=\frac{(1-x)^n-1}{x}
=-\sum^n_{j=1}(1-x)^{j-1}$$

$$\int^1_0\sum_{j=1}^n {n \choose j}(-1)^{j}x^{j-1}dx=-\sum^n_{j=1}\int^1_0(1-x)^{j-1}dx=-\sum^n_{j=1}\frac{1}{j}$$
$$\sum_{j=1}^n {n \choose j}\frac{(-1)^{j+1}}{j}=H_n$$

 

[Albert1]

Show that

$$\sum_{j=1}^{j=n}\binom{n}{j} \frac{(-1)^{j+1}}{j} = 1 +\frac{1}{2} +\frac{1}{3} + \cdots +\frac{1}{n}---(1)$$

Spoiler

to prove (1)we can also use the method of induction
it is easy to show that (1) is true when n=1,2,3--
suppose (1) is true as n=m ,now we need to prove :
$$\sum_{j=1}^{j=m+1}\binom{m+1}{j} \frac{(-1)^{j+1}}{j} = 1 +\frac{1}{2} +\frac{1}{3} + \cdots +\frac{1}{m+1}---(2)$$
the proof of (2):
we use the formula :

$$\binom{m+1}{j} =\binom{m}{j}+\binom{m}{m+1-j} $$

and the rest is not hard (many terms can be canceled)
now it is too late , I am going to sleep ,hope someone can finish it