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I have an infinite series $\sum\limits_{n=1}^{\infty}(1+\frac{2}{n})^n$. I need to show if it converges or diverges using any test. I've tried applying all of the tests that I know, and hit dead ends with each. I must have made a mistake somewhere but I can't figure it out. Here's the tests/steps I used but I'm not sure if I applied the rules correctly:

Divergence test: The test is inconclusive because $\lim\limits_{n\to\infty}\left(\frac{2+n}{n}\right)^n$ produces $1^\infty$ which is an indeterminate form.

Edit: The limit actually doesn't produce an indeterminate form, as pointed out in comments...so the divergence test shows that it diverges.

Geometric Series: I couldn't figure out how to apply this test because what would be "$r$" has $n$ in the denominator.

Integral test: I tried writing the integral $\int_1^\infty{\left(\frac{2+n}{n}\right)^n}dn$, but I'm not sure how to solve this because when I simplified the function inside by distributing the power to the numerator and denominator, I got $n^n$ which I'm not sure how to integrate.

Root test: When I applied this test, I got 1, which means the test is inconclusive.

Ratio test: When I applied this test, I wasn't sure how to go further after a few steps of simplification. $$\lim_{n\to\infty}\left(\frac{2+(n+1)}{n+1}\right)^{n+1}\cdot\left(\frac{n}{2+n}\right)^n$$

Some simplifying: $$\lim_{n\to\infty}\frac{(3+n)^{n+1}}{(n+1)^{n+1}}\cdot\frac{n^2}{(2+n)^n}$$

Here, I'm not sure how to simplify it further because the bases are different so I can't simply add the exponents, and I can't see that it simplifies any further. Plugging in infinity to the n's produces multiple indeterminante forms of $\infty^\infty$.

Perhaps I'm missing something simple, or I made a mistake applying the tests somewhere, but I'm stuck and not sure what I missed. Any help is appreciated.

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    $\begingroup$ Does the $n$-th term go to zero? There is this number, $e$. $\endgroup$ – Pedro Tamaroff Apr 11 '14 at 1:29
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    $\begingroup$ By the binomial theorem, $$\left(1+\frac{2}{n}\right)^n = \frac{(n+2)^n}{n^n}\geq\frac{n^n+2n\cdot n^{n-1}}{n^n}=3.$$ $\endgroup$ – Jack D'Aurizio Apr 11 '14 at 1:54
  • $\begingroup$ Remember this: $$ \lim_{n\rightarrow \infty} \left( 1+\frac{x}{n} \right)^n = e^x $$ $\endgroup$ – Dylan Apr 12 '14 at 22:25
  • $\begingroup$ The ratio test can work too, if you take the log of the limit, but the above is easier $\endgroup$ – Dylan Apr 12 '14 at 22:31
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Hint: Consider the limit of the sequence

$$a_n=\left(1+\frac{2}{n}\right)^n$$

You should find that this is nonzero, so that your series diverges.

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  • $\begingroup$ That was the first test I applied but taking the limit produces $1^\infty$, an indeterminate form. So even though it's indeterminate, it still counts as "non zero"? $\endgroup$ – Sabien Apr 11 '14 at 1:32
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    $\begingroup$ Well, it's not indeterminate, you can certainly find its limit to be $e^2$ (or at the very least to be greater than or equal to $1$) $\endgroup$ – MCT Apr 11 '14 at 1:33
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    $\begingroup$ Notice that $1+\frac{2}{n}>1$. When you raise any number greater than $1$ to a positive power, it is necessarily greater than $1$... $\endgroup$ – Cameron Williams Apr 11 '14 at 1:33
  • $\begingroup$ I see, I didn't think to transform the limit by writing $e^{(\lim)}$, that makes sense. Thanks! $\endgroup$ – Sabien Apr 11 '14 at 1:36
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    $\begingroup$ @Sabien: Do you know how to deal with indeterminate forms? We can use L'Hopital as is done in Chris's post, or we can use the fact that $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ $\endgroup$ – Jared Apr 11 '14 at 1:36
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Should be obvious that your series is bigger then $$\sum_{n=1}^\infty 1+e_n$$ for some $e_i> 0$ for all $i$, which is clearly divergent.

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What you want is to evaluate

$$\lim_{n \rightarrow \infty} (1+2/n)^n \\ = \exp(\lim n\cdot \log(1+2/n)) \\ = \exp(\lim \frac{-2/n^2}{-(1+2/n)/n^2}) \\ = \exp(2) \neq 0.$$

Note that a necessary condition for convergence would be that this limit goes to $0$.

But a less computationally-intensive method would be using the comparison test: compare each term with $1$.

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