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I have to find whether the infinite series $$\sum_{n=0}^\infty \left(\frac{n}{b(n-3)}\right)^n, \quad\text{for }b>0$$ diverges or find for which values of $b$ it converges to some finite value. So far I've tried to apply the ratio test but it didn't seem to work. Any advice on how to solve the problem is greatly appreciated.

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    $\begingroup$ Try the root test. $\endgroup$ – Gary Nov 25 '20 at 18:38
  • $\begingroup$ Your sum starts at n=0, hence if you don't exclude the term for n=3 then the sum i divergent trivially for any value of b, $\endgroup$ – Dr. Wolfgang Hintze Nov 25 '20 at 19:33
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You have $$\left(\frac{n}{b(n-3)}\right)^{n} = \frac{1}{b^n}\left(1-\frac{3}{n}\right)^{-n}$$

and $$\lim_{n \rightarrow +\infty} \left(1-\frac{3}{n}\right)^{-n} = e^3$$

so $$\left(\frac{n}{b(n-3)}\right)^{n} \sim \frac{e^3}{b^n}$$

So the series converges iff $$\frac{1}{b}<1$$ i.e. iff $b>1$.

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  • $\begingroup$ What does the tilde notation on line 3 mean? Does it mean that those two expressions are equal in the limit? $\endgroup$ – Vladimir Lenin Nov 25 '20 at 18:53
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    $\begingroup$ That means that the quotient of one by the other tends to $1$ : see en.wikipedia.org/wiki/Asymptotic_analysis (and don't worry if you don't know this notation yet, you will see when you learn it that it is very useful) $\endgroup$ – TheSilverDoe Nov 25 '20 at 18:56
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I'm using the root test, which you can read about if you want here: https://en.wikipedia.org/wiki/Root_test .

Consider $\lim \limits_{n\rightarrow \infty}((\frac{n}{b(n-3)})^n)^{\frac{1}{n}}=\lim \limits_{n\rightarrow \infty}\frac{n}{b(n-3)}=\lim \limits_{n\rightarrow \infty}\frac{1}{b}=\frac{1}{b} $ by L'hopital's rule. By the root test, the series converges if this limit is less than $1$, so the series converges if $b>1$.

Hope this helps.

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Actually, having all the expression affected by an n power, you will probably have better results using the root test. This is:

$L = \sqrt[n]{|{a_n}|} = \frac{n}{b(n-3)}$.

As that is a quotient of same-degree polynomials, the limit to infinity is:

$L = \lim_{n\to\infty} \sqrt[n]{|{a_n}|} = \frac{1}{b}$

As you may know, $L < 1$ ensures convergence, and $L > 1$ ensures divergence, thus the first clear conclusion is that if $b > 1$, the series converges, and if $b < 1$ the series diverges

The case $b = 1$ must be analysed separately, for in that case the root test does not give any information on the series' convergence. However, we can see that if $b = 1$:

$\lim_{n\to\infty} {a_n} = \lim_{n\to\infty} (\frac{n}{n-3})^n = \lim_{n\to\infty} (1 + \frac{3}{n-3})^n = e^3$

As $e^3 \neq 0$, the series is not convergent when $b = 1$, thus converging only when $b > 1$

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You can use the root test \begin{align*} \limsup_{n \to \infty} \sqrt[n]{\left( \frac{n}{b(n-3)} \right)^n} &= \limsup_{n \to \infty} \frac{n}{b(n-3)} \end{align*} The series converges, when $$ \limsup_{n \to \infty} \frac{n}{b(n-3)} > 1 $$ Therfore, for the series to converge, you have to choose $b$ as $$ b > \lim_{n\to \infty} \frac{n}{n-3} = 1 $$

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