# Show that the following series is convergent.

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.

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

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

• I am not sure that I can prove this inequality. Could you please elaborate? – nls Oct 30 '18 at 1:18
• If $0 < z \leqslant 1$ then $1 - z^m \leqslant m (1-z)$. Take $z = (S_{n-1}/S_n)^{1/m}$. This appears in the denominator and is $\leqslant 1$ Because the partial sums are nondecreasing. – RRL Oct 30 '18 at 1:22
• It took me a minute to see how that term "$m$" appears in the inequality........+1............ If I was fussy I would say it's all only valid when $a_n\ne 0$ (i.e. when $S_{n-1}/S_n\ne 1.$) But terms with $a_n=0$ can of course be ignored. – DanielWainfleet Oct 30 '18 at 1:45
• To the proposer: If $0\leq z\leq 1$ then $(1-z ) m \geq (1-z)\sum_{j=0}^{m-1}z^j=1-z^m.$ – DanielWainfleet Oct 30 '18 at 1:50
• @DanielWainfleet: Thanks for that helpful contribution to the answer. – RRL Oct 30 '18 at 4:36