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The sum is $$\sum_{n=1}^{\infty} \frac{n+2^n}{n+3^n}$$ Is this convergent or divergent? I tried to use the divergent test but the test fail because $a_n = (n+2^n)/(n+3^n) = 0 $ as $n$ goes to infinity. Could someone point me to the right direction?

Thanks

Bombyx mori
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afsdf dfsaf
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1 Answers1

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If you don't want to resort to the "big mallets" of the Ratio Test or the Limit Comparison Test, you could also note that

$$ \frac{n + 2^n}{n + 3^n} \ < \ \frac{2 \ \cdot \ 2^n}{n + 3^n} \ < \ \frac{2 \ \cdot \ 2^n}{ 3^n} \ , \ \text{for} \ n \ \ge \ 1 \ , $$

the last term in this inequality being the general term of a convergent geometric series.

colormegone
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  • This only show a_n converges to 0 as n goes to infinity. The divergence test fails because a_n =0 does not guarantee convergence or divergence of the sum of the series. – afsdf dfsaf Oct 31 '13 at 03:17
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    I am using the (simple) Comparison Test: your general term is smaller than the general term of a known convergent series. Thus, your series also converges. – colormegone Oct 31 '13 at 03:28