Thread: More on series converge or diverge?

Hi again, as I was doing more questions on this, I got stuck on two more questions.

1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
2. ∑(from n=1 to infinity) ( ln (1+1/n) )
Well Q1, same as before I tried ratio test and it turned out weird...
For Q2,not sure if I did it right, here's what I did. I compute the partial sums:
s1= ln(1+1)
s2= ln(1+1) + ln(1+1/2) > ln(1+1/2) + ln(1+1/2) = 2 ln(1+1/2)
and so lim for n to infinity for ln(1+1/n) = infinity
therefore the sequence of the partial sum diverges to infinity and so the sequence also diverges. Is this right?

Thanks.

Hello,

Originally Posted by Rui

Hi again, as I was doing more questions on this, I got stuck on two more questions.

1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )

You have to use the alternate series test, proving the absolute value of the summand is decreasing and that its limit when n tends to infinity is 0.
Or you can take a look at the alternate Riemann series, which state that converges if

Originally Posted by Rui

1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )

Use alternating series test:









So this satisfies the alternating series test, and is therefore convergent.

Originally Posted by Rui

2. ∑(from n=1 to infinity) ( ln (1+1/n) )

let

Then lets evaluate this as a partial sum, looking at the telescoping aspect.



cancel out common terms, and see that -ln(1) = 0 to get


so which clearly diverges.