Convergence by using Cauchy Criterion

Question

this is the sequence:

$(a_n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdot\cdot\cdot+\frac{1}{2n}$

And this is what I tried to do so far:

$|a_{n+1} - a_{n} | = \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1} = \frac{1}{2n+1}-\frac{1}{2n+2}$ in order to show (hopefully?) that the sequence IS convergent, since it is (hopefully) a Cauchy-Sequence.

However, as you can see, I don't know if a) I'm on the right way, b) what to do next.

score 1 · Answer 1 · answered May 14 '13 at 13:34

1

You have $${1\over n + k} \ge \int_{n+k}^{n+k+1}{dx\over x}.$$ Hence $$\sum_{k = 0}^n {1\over n + k} \ge \sum_{k=0}^n\int_{n+k}^{n+k+1}{dx\over x} = \int_{n}^{2n}{dx\over x} = \log(2).$$ Another similar inequality lurks. You can show this, in fact, converges, and that it is therefore Cauchy.

answered May 14 '13 at 13:34

ncmathsadist

49,383

hey, is there a second option without integral? – Vazrael May 14 '13 at 16:51
Likely, but it would involve some moderately arcane analytical trickery. I can't think of one off-hand. – ncmathsadist May 15 '13 at 23:57
how the first inequality you got? – Myshkin May 16 '13 at 13:07
1

Look at the graph of $x\mapsto 1/x$. – ncmathsadist May 16 '13 at 14:24

score 0 · Answer 2 · answered Oct 27 '15 at 11:19

0

Since $ 0 \le a_{n+1} - a_n , a_n \le 1 ( \forall n \in N ) $, use the Monotone Convergence Theorem.

answered Oct 27 '15 at 11:19

H.W.Kim

11

Convergence by using Cauchy Criterion

2 Answers2