Show that the following series is convergent.

Question

Consider the series whose general term is as follows: $$u_n=\frac{a_n}{(S_n)^\lambda}$$ with the condition $S_n = \sum_{k=1}^{n}a_k$ with constraints that $0\leq a_n\leq 1,$ $S_n$ is a divergent series and $\lambda >1.$ Show that the series is convergent.

I need to find a lower bound for $S_n$ so that I can find an upper bound for $u_n.$ I tried to use the fact that $S_n$ is divergent in the following way:

For $n$ large enough we can say that $S_n>N$ where $N>1$ and but this gives the bound $$u_n<\frac{1}{N^\lambda}$$ which is not helpful since we will sum up constant terms infinite times. Any hints/suggestions will be much appreciated.

Answer 1

Note that with $\lambda > 1$ there is an integer $m$ such that $\frac{1}{m} < \lambda - 1$ and for $n > 1$

$$\tag{*}\frac{a_n}{S_n^\lambda} \leqslant \frac{a_n}{S_n S_{n-1}^{\lambda-1}} \leqslant \frac{S_n - S_{n-1}}{S_n S_{n-1}^{1/m}} = \frac{1- \frac{S_{n-1}}{S_n}}{1- \frac{S_{n-1}^{1/m}}{S_n^{1/m}}}\left(\frac{1}{S_{n-1}^{1/m}} - \frac{1}{S_n^{1/m}} \right) \\ \leqslant m\left(\frac{1}{S_{n-1}^{1/m}} - \frac{1}{S_n^{1/m}} \right)$$

The sum of the term on the RHS of (*) is telescoping and converges since $1/S_n \to 0$:

$$\sum_{n=2}^\infty m\left(\frac{1}{S_{n-1}^{1/m}} - \frac{1}{S_n^{1/m}} \right) = \frac{m}{S_1^{1/m}} $$

By the comparison test $\sum a_n/S_n^\lambda $ converges.

See if you can prove the far right inequality in (*), that is

$$\frac{1- \frac{S_{n-1}}{S_n}}{1- \frac{S_{n-1}^{1/m}}{S_n^{1/m}}} \leqslant m$$