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Say I have this series:

$$\sum_{n=1}^{\infty} \frac{n+3^n}{n+2^n}$$

I think I can compare it to $$b = \frac{3}{2}^n$$ which is a geometric series that diverges.

But where do I go from here? The issue is that the numerator and denominator both increase so I can't say that the original series diverges or not by Comparison Test. What else can I do?

Jwan622
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2 Answers2

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As Lord Shark the Unknown commented, since the numerator grows larger quicker than the denominator, we can simply use the divergence test to conclude that the series diverges. $$\lim_{n\to \infty} \frac{n+3^n}{n+2^n} = \infty \neq 0 \Rightarrow \text{diverges}$$

A comparison test is also appropriate. As you noticed, you can compare your sequence to $\left(\frac{3}{2}\right)^n$. I would recommend a limit comparison test: $$\lim_{n\to \infty} \frac{n+3^n}{n+2^n} \cdot \frac{2^n}{3^n} =\lim_{n\to \infty}\frac{n+3^n}{n+2^n}\cdot \frac{\frac{1}{3^n}}{\frac{1}{2^n}}= \lim_{n \to \infty} \frac{\frac{n}{3^n}+1}{\frac{n}{2^n}+1} =1$$ Since $0<1<\infty$, and $\sum\left(\frac{3}{2}\right)^n$ diverges, the original sum has the same behavior and diverges as well.

BSplitter
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  • How do I formally show taht the limit approaches infinity? – Jwan622 Apr 03 '19 at 03:49
  • One way to do that is to multiply both the numerator and denominator by $\frac{1}{2^n}$. Then you get the limit: $$\lim_{n \to \infty} \frac{n/2^n + (3/2)^n}{n/2^n+1} $$ Although it's not notationally correct, we could "plug in infinity" and get $\frac{0+\infty}{0+1} = \frac{\infty}{1} = \infty$ – BSplitter Apr 03 '19 at 03:53
  • It's an axiom that n/2^n goes to 0 as n -> infinity right? No further proof needed? How did you do the multiplication to get the new limit in the limit comparison test? I think I'm missing a step. – Jwan622 Apr 03 '19 at 03:54
  • We wouldn't call it an axiom, but yes, there are theorems that state that $\lim_{n\to \infty} \frac{n}{2^n} = 0$. Most likely, no further work would be necessary.

    As for the LCT, I'm computing $\lim_{n\to \infty} \frac{a_n}{b_n}$, where $a_n = \frac{n+3^n}{n+2^n}$ and $b_n = \frac{3^n}{2^n}$.

     – BSplitter Apr 03 '19 at 03:58
  • How did you get to $$lim_{n \to \infty} \frac{\frac{n}{3^n} + 1}{\frac{n}{2^n} + 1}$$ – Jwan622 Apr 03 '19 at 04:00
  • I edited my answer. Does that clear up the confusion? – BSplitter Apr 03 '19 at 04:01
  • Regarding $n/2^n \to 0$, see https://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial (and its many duplicates). – Hans Lundmark Apr 03 '19 at 04:35
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One of my sayings is that you don't need to get the best possible values when finding limits - good enough is good enough.

In this case $\frac{n+3^n}{n+2^n} \gt \frac{3^n}{2\cdot 2^n} =\frac12(\frac32)^n $ and the sum of these clearly diverges - heck, the individual terms diverge.

marty cohen
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