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http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf

In problem (2), at the very end it says

$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$

I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$?

Is one allowed to do the following?

$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$

$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$

$$(0 + 0 + 0 + \dots)^{1/2} = 0$$

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    $\begingroup$ is it $\frac{1}{k^2}$? $\endgroup$ – i.a.m Feb 22 '13 at 21:31
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    $\begingroup$ If a series converges then the "sum of its tail" goes to zero. If $\sum_{n=1} a_n$ converges then $\sum_{n>K} a_n$ goes to zero as $K$ goes to infinity. $\endgroup$ – Maesumi Feb 22 '13 at 21:33
  • $\begingroup$ Your limit argument is not correct. Such an approach only applies to finitely many terms. If you are applying a limit to a sum of a few things then you can take the limits individually and add them if they exist. But when you have limit of a sum of infinitely many terms you cannot take the limit of individual terms and then add them later. When you study Riemann sum you see a clear example of this. $\endgroup$ – Maesumi Feb 22 '13 at 21:38
  • $\begingroup$ This really isn't a series. It's a sequence, and part of the definition of each term is a series. But it's important to understand that you really have two limits here, one as $n$ goes to $\infty$, but another as $k$ goes to $\infty$, which is evaluated for each $n$. $\endgroup$ – MartianInvader Feb 22 '13 at 22:47
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Feb 22 '13 at 23:39
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One can forget about the square root part for a while. Note that $\dfrac{1}{k^2}\lt \dfrac{1}{(k-1)k}$.

But $\dfrac{1}{(k-1)k}=\dfrac{1}{k-1}-\dfrac{1}{k}$. Thus your sum is less than $$\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}+\cdots.$$ Note the wholesale cancellation: the above sum is $\dfrac{1}{n}$.

It follows that your original expression is less than $\dfrac{1}{\sqrt{n}}$.

Remark: Treating infinite "sums" as if they were long finite sums is a dangerous business that can all too easily give wrong answers. If one has experience with a particular series, such as the convergent series $\sum_1^\infty \frac{1}{n^2}$, then one can "see" that the tail must approach $0$. In fact, the issue is precisely the issue of the convergence of $\sum_1^\infty \frac{1}{n^2}$.

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It turns out that

$$\sum_{k=N+1}^{\infty} \frac{1}{k^2} = \int_0^{\infty} dx \frac{x }{e^x-1} e^{-N x}$$

You can prove this by factoring out an $e^{x}$ from the denominator and Taylor expanding the resulting denominator. In any case, by integrating by parts, you can show that

$$\sum_{k=N+1}^{\infty} \frac{1}{k^2} = \frac{1}{N} + O\left( \frac{1}{N^2}\right)$$

In fact, you can get a full asymptotic expansion of the sum over $N$ using this integral. In any case, though, this shows how the sum vanishes as $N \rightarrow \infty$.

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    $\begingroup$ I think this answer may be better for the OP than the more abstract answers. $\endgroup$ – GEdgar Feb 22 '13 at 22:16
  • $\begingroup$ Thanks. Actually, this is a result I first saw derived in Bender & Orszag, and I think is a little pearl more people should have in their toolbox. $\endgroup$ – Ron Gordon Feb 22 '13 at 22:17
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First note that $$\dfrac1{k^2} \leq \int_{k-1}^k\dfrac{dx}{x^2} \,\,\,\, (\text{Why?})$$ Hence, $$\sum_{k=n+1}^{\infty} \dfrac1{k^2} \leq \sum_{k=n+1}^{\infty} \int_{k-1}^k\dfrac{dx}{x^2} = \int_n ^{\infty} \dfrac{dx}{x^2} = \dfrac1n$$ Hence, we have that $$0 \leq \lim_{n \to \infty} \sum_{k=n+1}^{\infty} \dfrac1{k^2} \leq \lim_{n \to \infty} \dfrac1n = 0$$ Your argument to prove it is incorrect, since in general you cannot swap two limits to conclude the answer. For instance, by your same argument, you will also get that $$\lim_{n \to \infty} \left(\dfrac1{n} + \dfrac1{n} + \cdots + \dfrac1n + \cdots \right) = 0$$which is clearly false.

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In general we have if $\sum_{k=1}^{\infty}a_k$ converges then the tail of the sereis $\sum_{k=N}^{\infty}a_k$must go to zero to see this consider $$s_n=\displaystyle\sum_{k=1}^{n}a_k$$ then we have$s_n\to\sum_{k=1}^{\infty}a_k$ that means for any $\epsilon$ we there is $N$ such that$$|\displaystyle\sum_{k=1}^{\infty}a_k-s_N|<\epsilon$$ but $$|\displaystyle\sum_{k=1}^{\infty}a_k-s_N|=|\displaystyle\sum_{k=N}^{\infty}a_k|$$ thus, $$|\displaystyle\sum_{k=N}^{\infty}a_k|<\epsilon$$ since epsilon was arbitrary we conclude that$\sum_{k=N}^{\infty}a_k\to 0$.

as for your example we know by the integral test that $\sum_{k=1}^{\infty}\frac{1}{k^2}$ is convergent Thus, by what I wrote we must have $\sum_{k=N}^{\infty}\frac{1}{k^2}\to 0.$

Remark: of course if $\sum_{k=N}^{\infty}\frac{1}{k^2}\to 0.$ then $(\sum_{k=N}^{\infty}\frac{1}{k^2})^{1/2}\to 0.$

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Since the sum $\sum_{k=1}^\infty \frac{1}{k^2}$ is convergent, you can make the "tail" of the sum ($\sum_{k=n+1}^\infty \frac{1}{k^2}$) as small as you want.

Explicitly, since $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$, for any $\epsilon>0$, you can find an $n$ such that $\sum_{k=n+1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}-\sum_{k=1}^n \frac{1}{k^2}<\epsilon$.

In that case, the tail sums form a strictly decreasing sequence of positive numbers, which can be made smaller than any $\epsilon>0$.

That is convergence to 0! :)

Naturally if the expression converges to zero, so does the square root of the expression.

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  • $\begingroup$ $\sum_{k=n+1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}-\sum_{k=1}^n \frac{1}{k^2}$ Are you sure you can break up the sum like that? $n$ is a variable here no? $\endgroup$ – Hawk Feb 22 '13 at 22:19
  • $\begingroup$ @sizz I don't see why $n$ being a variable is scary: it's true for every $n$. The thing on the left hand side is an absolutely convergent series, and it's clearly equal to the right hand side. $\endgroup$ – rschwieb Feb 23 '13 at 1:07
  • $\begingroup$ @sizz in fact you can see two other duplicates of my answer already :/ $\endgroup$ – rschwieb Feb 23 '13 at 1:09
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Notice that $\sum \frac{1}{k^2}$ is convergent. If I invoke Cauchy's Convergence Criterion, the proof of this sum is trivial.

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A related problem. Note that,

$$ \sum_{k=n+1}^{\infty} \frac{1}{k^2}= \sum_{k=1}^{\infty} \frac{1}{(k+n)^2} \leq \sum_{k=1}^{\infty} \frac{1}{k^2} <\infty, $$

which implies that the series

$$ \sum_{k=1}^{\infty} \frac{1}{(k+n)^2} $$

converges uniformly. So, we have

$$ \lim_{n\to \infty} \sqrt{ \sum_{k=n+1}^{\infty}\frac{1}{k^2} } = \sqrt{\lim_{n\to \infty} \sum_{k=n+1}^{\infty}\frac{1}{k^2} } = \sqrt{\lim_{n\to \infty} \sum_{k=1}^{\infty}\frac{1}{(k+n)^2} } = \sqrt{ \sum_{k=1}^{\infty}\lim_{n\to \infty}\frac{1}{(k+n)^2} }=0. $$

Notice that, the function $\sqrt{x}$ is a continuous function for $x>0$, which justifies changing the operation $\lim f(a_n) = f( \lim a_n )$.

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