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The question:
Test the following series for convergence or divergence: $$ \frac{1}{2^2}+\frac{\sqrt{2}}{3^2}+\frac{\sqrt{3}}{4^2}+\frac{\sqrt{4}}{5^2}+\dots $$ My solution
The general function would be $$\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{k}}{(k+1)^2}$$ so $$\lim_{n\rightarrow\infty}\sqrt{n}=\infty$$ also $$\lim_{n\rightarrow\infty}(n+1)^2=\infty$$ so the series is convergent because $\frac{\infty}{\infty}=1$, no?

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It does converge, but not because $\frac{\infty}{\infty}=1$ (which if meaningful would suggest divergence when summed, but it is not meaningful) –  Henry Nov 20 '12 at 13:45
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Note, you use $\sum_{k=1}^n$ but use $n$ inside. You should write: $$\sum_{k=1}^n\frac{\sqrt{k}}{(k+1)^2}$$ –  Thomas Andrews Nov 20 '12 at 13:47
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Minor comment: One says $\sum_{k=1}^\infty a_k$ converges, or $\lim_{n\to \infty}\sum_{k=1}^n a_k$ exists, not the mixed expression in the title. –  André Nicolas Nov 20 '12 at 17:11
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6 Answers 6

up vote 5 down vote accepted

Simply saying $\frac\infty\infty$ is meaningless. Does $\sum_{n=1}^\infty\frac{n}{n}$ converge? No. In this case, $$\lim_{n\to\infty}\frac{\sqrt{n}}{(n+1)^2}=0,$$ which is very important, since if these terms don't tend to $0$, then the series can't possibly converge.

In this case, you'll want to use direct comparison test and integral test, instead. $$0<\frac{\sqrt{n}}{(n+1)^2}<\frac{\sqrt{n+1}}{(n+1)^2}=\frac1{(n+1)^{3/2}},$$ So since $$\int_1^\infty\frac1{(x+1)^{3/2}}\,dx$$ converges (check), then $$\sum_{n=1}^\infty\frac1{(n+1)^{3/2}}$$ converges by the integral test, and so $$\sum_{n=1}^\infty\frac{\sqrt{n}}{(n+1)^2}$$ converges by direct comparison test.

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No.

First, "$\infty/\infty=1$'' makes no sense.

Second, even if it did, one could apply your argument to show that $\sum_{k=1}^\infty \frac{n}n$ converges.

The series you are considering converges because it is comparable with a convergent series. Namely, $$ \frac{\sqrt n}{(n+1)^2}\leq\frac{\sqrt n}{n^2}=\frac1{n^{3/2}}. $$ As $\sum_{k=1}^\infty n^{-3/2}<\infty$, we conclude by comparison that $\sum_{k=1}^\infty \sqrt{n}/(n+1)^2<\infty$.

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The series indeed converges. Since $(n+1)^2 > n^2$, we have

$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)^2} < \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2} = \sum_{n=1}^{\infty} n^{\frac{-3}{2}}$.

Now it is known by integral test that $\sum_{n=1}^{\infty}\frac{1}{n^s}$ converges for $s>1 \in \mathbb{R}$.

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$$0 \lt \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{n}}{(n+1)^2} \lt \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{n+1}}{(n+1)^2} \lt \lim_{n\rightarrow\infty}\int_{x=1}^{n+1} x^{-3/2} \, dx = 2. $$ So the series limit converges on a number less than $2$ from below.

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In fact less than $1.421$ –  Henry Nov 20 '12 at 14:00
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You can use Limit comparison test. Since $$\lim_{n \to \infty}\dfrac{\frac{\sqrt{n}}{(n+1)^2}}{\frac{1}{n^\frac{3}{2}}}=1$$ and the series $\displaystyle{\sum_{k=1}^\infty\frac{1}{n^\frac{3}{2}}}$ converges, the series $\displaystyle{\sum_{k=1}^\infty\frac{\sqrt{n}}{(n+1)^2}}$ converges.

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In short: yes, but not for that reason. What you are doing is not a valid test for convergence. For example, $a_n = \frac{n^2}{n}$ both top and bottom converge to $\infty$ yet the series certainly diverges. Any argument involving "$\frac{\infty}{\infty} = 1$" is likely to be wrong.

However, the series does converge for several reasons. I like $a_n < \frac{\sqrt{n+1}}{(n+1)^2} = (n+1)^{-\frac{3}{2}}$ which converges since $\frac{3}{2} >1$, a well known result (which can be proved by the integral test)

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