# Divergence of $\lim \limits_{t \to \infty}\sum_{n=1}^t b^{\ln n}$ and the Monotonic Sequence Theorem

As far as I understand, the Monotonic Sequence Theorem states that if a sequence is monotonic and the individual terms are bounded, then the sequence is convergent.

My book states that $$\lim \limits_{t \to \infty}\sum_{n=1}^t b^{\ln n}$$ is convergent only for b $$\lt$$ $$\frac{1}{e}$$. However, is $$\lim \limits_{t \to \infty}\sum_{n=1}^t 0.5^{\ln n}$$, for example, not a monotonically decreasing series, whose terms are bound by 0 below and 1 above? Therefore this series meets the criteria for the MST, yet diverges, and does not share the outcome predicted by the theorem.

Why is $$\lim \limits_{t \to \infty}\sum_{n=1}^t 0.5^{\ln n}$$ divergent when it appears to meet the Monotonic Sequence Theorem's criteria for convergence?

• There is a difference between the sequence and the series. The sequence ${1 \over n}$ is bounded and monotonic and has a limit, but $\sum_k {1 \over k}$ does not. – copper.hat Jun 23 at 18:30

You've confused sequence convergence (the $$n$$th term has a finite $$n\to\infty$$ limit) with series convergence (the sum of the first $$n$$ terms has a finite $$n\to\infty$$ limit).
$$\lim_{t\to\infty} \sum_{n=1}^\infty b^{\ln n}=\sum_{n=1}^\infty b^{\ln n}=\sum_{n=1}^\infty e^{\ln b\ln n}=\sum_{n=1}^\infty n^{\ln b}$$ This series converges iff $$\ln b<-1$$,i.e. when $$b<1/e$$.
Now, $$0.5>1/e$$. So, you certainly have that when $$b=0.5$$ this diverges. This doesn't contradict monotonic sequence theorem, even though $$(0.5)^{\ln n}$$ is bounded by $$1$$ for all $$n$$, you don't have that $$\sum_{n=1}^\infty (0.5)^{\ln n}$$ is bounded for all $$t$$.