5
$\begingroup$

Let $u_n$ be a sequence of positive numbers, for each n: ${u_{n+1}\over{u_n}}\le(\frac{n}{{n+1}})^\alpha$ when $\alpha>1$. Prove that $\sum_{n=1}^\infty {u_n}$ converges.

I would like to get a hint.

$\endgroup$
2
$\begingroup$

Hint: $${u_{n+1}}\le\left(\frac{n}{{n+1}} \right)^\alpha u_n$$ So we get $$u_n \le \frac{u_1}{n^\alpha},$$ and apply comparison test.

$\endgroup$
0
$\begingroup$

The hypothesis implies that the sequence $n^\alpha u_n$ is non-increasing. In particular, it is bounded by $M = u_1$. Since $\alpha > 1$, this yields $$ \sum_{n=1}^\infty u_n \leq \sum_{n=1}^\infty\frac{M}{n^\alpha} < \infty $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.