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I need to test whether the following series is convergent or not: $$\sum_{n=1}^{\infty} \left(\frac{2}{2n+1}\right)^n$$

I thought of using the Limit Comparison Test with a geometric series $b_n=\left(\frac{1}{2}\right)^n$. If $a_n = \left(\frac{2}{2n+1}\right)^n$, then I can create the following limit: $$\lim_{x \to \infty} \frac{\left(\frac{2}{2n+1}\right)^n}{\left(\frac{1}{2}\right)^n} = \lim_{x \to \infty} \left(\frac{4}{2n+1}\right)^n$$

The limit approaches $0^{\infty}$ so I decide to use logarithms.

$$\ln{y} = \lim_{x \to \infty} n*\ln\left(\frac{4}{2n+1}\right)$$ This limit approaches infinity if I try to use logarithms, but $\left(\frac{2}{2n+1}\right)^n$ is a convergent series. Am I using the wrong test or is the way I'm calculating my limit incorrect?

EDITED WORK

1: Ratio Test$$\lim_{x \to \infty} \frac{(\frac{2}{2n+1})^n}{(\frac{2}{2n-1})^{n-1}} = 2 \lim_{x \to \infty} \frac{(2n-1)^{n-1}}{(2n+1)^n}$$

I am thinking of using L'Hopital's, but it will be cyclical with the derivatives.

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4 Answers 4

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Yes you are right,the given series is convergent You are using wrong$b_n$

Hint: Try$$b_n=\frac{1}{n^2}$$

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  • $\begingroup$ By the way ,mate how you're claiming that$$\lim_{n\to \infty} n*ln(\frac{4}{2n+1}) \rightarrow \infty$$? By using L'Hospital's rule?? $\endgroup$
    – Nimu Basak
    Commented Apr 26, 2020 at 18:52
  • $\begingroup$ Sorry I realized that my question had a typo on it. It's actually $\sum_{n=1}^{\infty} (\frac{2}{2n+1})^n$. $\endgroup$
    – geo_freak
    Commented Apr 26, 2020 at 18:56
  • $\begingroup$ I graphed $ln(\frac{4}{2n+1})$ and saw that it's approaching -$\infty$. $\endgroup$
    – geo_freak
    Commented Apr 26, 2020 at 18:57
  • $\begingroup$ Yes indeed it is.. Because as $n\rightarrow \infty \frac{4}{2n+1}\rightarrow 0\implies \ln(\frac{4}{2n+1})\rightarrow (-\infty)$but I was wondering about that $$lim_{n\to \infty} n*ln(\frac{4}{2n+1}) \rightarrow \infty$$ where does it run.. $\endgroup$
    – Nimu Basak
    Commented Apr 26, 2020 at 19:05
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Apply the root test instead:$$\lim_{n\to\infty}\sqrt[n]{\left(\frac2{2n+1}\right)^n}=\lim_{n\to\infty}\frac2{2n+1}=0.$$

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  • $\begingroup$ Apologies, I realized that my series is $\sum_{n=1}^{\infty} (\frac{2}{2n+1})^n$. $\endgroup$
    – geo_freak
    Commented Apr 26, 2020 at 18:56
  • $\begingroup$ I've edited my answer. $\endgroup$ Commented Apr 26, 2020 at 19:02
  • $\begingroup$ I've added my ratio test work to my question, but my limit isn't working. $\endgroup$
    – geo_freak
    Commented Apr 26, 2020 at 19:12
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You could have used the ratio test $$a_n = \left(\frac{2}{2n+1}\right)^n\implies \log(a_n)=n\log\left(\frac{2}{2n+1}\right)$$ Using Taylor series $$\log(a_n)=-\log (n)-\frac{1}{2 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor series to get $$\log(a_{n+1})-\log(a_n)=-1-\log (n)-\frac{1}{2 n}+\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{a_{n+1} } {a_{n} }=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{e n}-\frac{1}{2 e n^2}+O\left(\frac{1}{n^3}\right)$$

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Let, $a_n=\left(\frac{2}{2n+1}\right)^n\ \forall n\in\Bbb{N}$. $a_n>0\ \forall n$. Now, apply root test $$\lim a_n^{1\over n}=\lim \frac{2}{2n+1}=0<1$$ So, by root test the series $\sum a_n=\sum \left(\frac{2}{2n+1}\right)^n$ converges.

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