Question

HELP PLEASE

Answer 1

Determine whether the series is convergent or divergent.
\[\sum_{n=1}^{\infty} n e^{-4n}\]

Answer 2

@zepdrix @mathmale

Answer 3

With these types of problems,
it's good to have some "idea" of what you think it will do.

That will give you some indication of what approach might be appropriate.

Convergent? Divergent? Any guess to start?

Answer 4

Ratio Test works well for this one,
Integral Test would be fine also.

Answer 5

Theorem (common sense really):
\(\color{blue}{\displaystyle \forall x\in \mathbb{N}{\rm{\small ~~~} and }{\small ~~~}\forall \alpha\in\mathbb{R}{\rm{\small ~~~} such~~that }{\small ~~~}|\alpha|>1,~S=\sum_{n=k}^{\infty}\frac{n^x}{\alpha ^n} {\small ~~~}{\rm converges}}\).
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Let \(\color{black}{\displaystyle 1<|\beta|<|\alpha| }\) for some \(\color{black}{\displaystyle \beta\in\mathbb{R} }\).
Then, \(\color{black}{\displaystyle \lim_{n\to\infty}\frac{n^x}{\beta^n}=0}\), and therefore (for all \(\color{black}{\displaystyle x\in\mathbb{N} }\)) there exists \(\color{black}{\displaystyle p\in\mathbb{N}}\),
such that \(\color{black}{\displaystyle n=k+p}\) satisfies \(\color{black}{\displaystyle \left|\frac{n^x}{\alpha^n}\right|<\left|\frac{\beta^n}{\alpha^n}\right| }\).
\(\tiny \\[1.2em]\)
Notice that the series \(\color{black}{\displaystyle \sum_{n=k+p}^\infty \frac{\beta^n}{\alpha^n}}\) converges because \(\color{black}{\displaystyle |r|=|\beta/\alpha|<1 }\),
and since \(\color{black}{\displaystyle \left|\frac{\beta^n}{\alpha^n}\right|>\left|\frac{n^x}{\alpha^n}\right|}\) for all \(\color{black}{\displaystyle n\ge k+p}\), therefore \(\color{black}{\displaystyle \sum_{n=k+p}^\infty \frac{n^x}{\alpha^n}}\) converges.
\(\tiny \\[1.2em]\)
Since \(\color{black}{\displaystyle \sum_{n=k+p}^\infty \frac{n^x}{\alpha^n}}\) converges, then also \(\color{black}{\displaystyle \sum_{n=k+p}^\infty \frac{n^x}{\alpha^n}}\) converges. \(\color{black}{\displaystyle ☐}\)

Answer 6

\(\color{red}{\bf Correction:}\)

The very last series is supposed to start from \(\color{red}{n=k}{\tiny~}\).

Answer 7

So, by this common sense theorem (which is easily provable - as I showed),
you will have:

\(\color{black}{\displaystyle \sum_{n=1}^\infty ne^{-4n}=\sum_{n=1}^\infty \frac{n}{e^{4n}}=\sum_{n=1}^\infty \frac{n^1}{(e^4)^n}}\)

(note that \(1\in\mathbb{N}\), and \(|e^4|>1\).)

Answer 8

To me, just change the "inside" to

\(\dfrac{4n}{4*e^{4n}}\),


Now you have the form \(\dfrac{1}{4}*lim_{4n\rightarrow \infty }\dfrac{4n}{e^{4n}}=\dfrac{1}{4}\)

Answer 9

@Loser66 that limit is 0.

Answer 10

yes, my bad. :)