Questions on Convergence of an Infinite series

Question

If the sequence of partial sums of an infinite series $$\sum_{n=1}^∞ a_n$$ is bounded, show that $$\sum_{n=1}^∞ a_ne^{-nt}$$ is convergent for $t>0$

Request: Actually I don't need a direct answer, I'm studying infinite series since the last couple of days. So I don't fully understand this thing. That's why I want to understand the process to solve this type of problems.

Thank you.

Answer 1

Dirichlet's test, mentioned in a comment by user49640, implies that the result is the same if $e^{-nt}$ is replaced by $b_n$ with $b_n$ monotonically decreasing to $0$.

Here there is a more elementary solution, and it applies if $e^{-nt}$ is replaced by $b_n$ with $b_n>0$ and $\sum\limits_{n=1}^\infty b_n$ convergent.

Because the partial sums of $\sum a_n$ are bounded, the sequence $(a_n)$ is also bounded, because if $\left|\sum\limits_{n=1}^N a_n\right|\leq M$ for all $N$, then $|a_N|=\left|\sum\limits_{n=1}^N a_n-\sum\limits_{n=1}^{N-1} a_n\right|\leq 2M$ by the triangle inequality. Therefore for each $N$, $\sum\limits_{n=1}^N |a_nb_n|\leq 2M\sum\limits_{n=1}^N b_n\leq 2M\sum\limits_{n=1}^\infty b_n<\infty$. By the monotone convergence theorem, $\sum a_n b_n$ is absolutely convergent, hence convergent.

Note that $\sum e^{-nt}=\sum (e^{-t})^n$ is convergent when $t>0$ by the geometric series test.

Answer 2

Actually the hypothesis are telling quite clearly which technique to use, i.e. summation by parts.
Let $A_n=\sum_{k=1}^{n}a_k$ and assume $N>1$. We have:

$$ \sum_{n=1}^{N} a_n e^{-nt} = A_N e^{-N t}+\sum_{n=1}^{N-1}A_n\left(e^{-nt}-e^{-(n+1)t}\right) $$ hence if $|A_n|\leq C$ for any $n\geq 1$, we have $$ \left|\sum_{n=1}^{N}a_n e^{-nt}\right|\leq \frac{C}{e^{Nt}}+ C(1-e^{-t})\sum_{n=1}^{N-1}e^{-nt}\leq C e^{-Nt} + C e^{-t} $$ and the given series is convergent.

Answer 3

Since the partial sums $S_n$ are bounded, so is the $n$-th term $a_n=S_n-S_{n-1}$. Therefore the series is absolutely convergent by comparision with a geometric series with ratio $e^{-t}$.

Answer 4

Abel theorem:

$$\sum^{\infty}_{n=1}a_{n}x_{n}$$ comply that: $$\left. \exists k\in>0/\forall n\in \mathbb{N}, \left| \sum^{n}_{j=1}a_{n}\right| <K\atop \lim_{n\to \infty}=0; x_{n}\geq x_{n+1}, \forall n\in \mathbb{N} \right\}\rightarrow \sum^{\infty}_{n=1}a_{n}x_{n}$$ is convergent. $$t\geq 0\rightarrow x_{n}=e^{-nt}\rightarrow \left. \lim_{n\to \infty}e^{-nt}=0\atop e^{-nt}=\dfrac{1}{e^{nt}}>\dfrac{1}{e^{(n+1)t}}=e^{-(n+1)t} \right\}$$ The fact that the partial sums of $\{a_{n}\}$ are bounded means that the sequence is small enough that, when multiplied by an infinite decreasing, we obtain another infinitesimal that gives rise to a convergent series.