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I am really struggling with coming up with comparison series for doing the Limit Comparison tests, especially when $n$ is in the exponent.

For the series $$\sum_{n=0}^\infty \frac{n +2^n}{n+3^n} $$

Could some walk me through how to find a comparison series for this question to determine convergence or divergence? I can only use the comparison test and not another test.

I want to compare it to $$\sum_{n=0}^\infty = \left(\frac{2}{3}\right)^n $$ Which is a convergent geometric series with $|r| = \frac{2}{3}<1$, but this answer is wrong and I am stumped.

user
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Sam
  • 865

3 Answers3

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As this is a series with positive terms, you can find a series with asymptotically equivalent terms to determine whether it converges or diverges.

Now $\;n+2^n\sim_\infty 2^n$, and similarly$\;n+3^n\sim_\infty 3^n$ , so $$\frac{n+2^n}{n+3^n}\sim_\infty\frac{2^n}{3^n}=\Bigl(\frac23\Bigr)^n,$$ which a convergent geometric series.

Bernard
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It can be done with two comparison series. For some $0 < t < 1$ we can take $$ \sum_{k=0}^\infty k t^k = \frac{t}{(1-t)^2} $$ This takes care, with $t=1/3,$ of $$ \frac{n}{n + 3^n} \leq \frac{n}{3^n} $$ Next you have your $$ \frac{2^n}{n + 3^n} \leq \frac{2^n}{3^n} $$

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well, why not. $ 0 < t < 1.$ $$ 1 + t + t^2 + t^3 + t^4 + \cdots = \frac{1}{1-t} $$ Derivative $$ 1 + 2t + 3 t^2 + 4 t^3 + \cdots = \frac{1}{(1-t)^2} $$ multiply by a single $t$ $$ t + 2 t^2 + 3 t^3 + 4 t^4 + \cdots = \frac{t}{(1-t)^2} $$

Will Jagy
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Yor solution is correct, indeed

$$\sum_{n=0}^\infty \left(\frac{2}{3}\right)^n$$

is a convergent geometric series and

$$ \frac{n +2^n}{n+3^n}\sim \left(\frac{2}{3}\right)^n$$

and therefore by limit comparison test the givan series converges.

user
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