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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?

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    $\begingroup$ 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. $\endgroup$ – copper.hat Jun 23 at 18:30
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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).

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$$\lim_{t\to\infty} \sum_{n=1}^\infty b^{\ln n}=\sum_{n=1}^\infty b^{\ln n}=\sum_{n=1}^\infty e^{\ln b\ln n}=\sum_{n=1}^\infty n^{\ln b}$$ 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 $t$.

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