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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.

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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

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

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  • $\begingroup$ I am not sure that I can prove this inequality. Could you please elaborate? $\endgroup$ – nls Oct 30 '18 at 1:18
  • $\begingroup$ 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. $\endgroup$ – RRL Oct 30 '18 at 1:22
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Oct 30 '18 at 1:45
  • $\begingroup$ To the proposer: If $0\leq z\leq 1$ then $(1-z ) m \geq (1-z)\sum_{j=0}^{m-1}z^j=1-z^m.$ $\endgroup$ – DanielWainfleet Oct 30 '18 at 1:50
  • $\begingroup$ @DanielWainfleet: Thanks for that helpful contribution to the answer. $\endgroup$ – RRL Oct 30 '18 at 4:36

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