I was wondering how to prove that a sequence is converging to a finite limit.
I know that if a sequence is bounded and decreasing / increasing then it has a finite limit.
although I have 2 questions where I don't know how to prove them :
the sequence : $a_n = \frac {1}{n} + \frac {1}{n+1} + … \frac {1}{3n}$
This sequence is bounded between 1 and 3 , I tried to prove its decreasing but failed to.
the sequence : $a_n = \frac {1}{\sqrt{1}} + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} + …+ \frac {1}{\sqrt{n}}$
is not bounded and has no finite limit ($\infty$)
what is the way to prove?
Thank you.
Best Answer
hint
$$a_n=\frac 1n+\frac{1}{n+1}+...+\frac{1}{3n}$$
$$a_{n+1}=\frac{1}{n+1}+...+\frac{1}{3n}+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3(n+1)}$$
then
$$a_{n+1}-a_n=$$ $$\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}-\frac 1n=$$ $$\frac{.......}{n(3n+1)(3n+2)(3n+3)}$$
For the second sequence, use the fact that for any $ k\in\{1,2,...n\}$,
$$ \frac{1}{\sqrt{k}}\ge \frac{1}{\sqrt{n}}$$ to get $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\ge \frac{n}{\sqrt{n}}$$
Observe that the RHS goes to infinity and conclude.