# Analysis, infinite series

• Oct 27th 2008, 12:56 PM
Maxmw22
Analysis, infinite series
OK, questions is about divergent, conditionally convergent, or abs convergent.

$
\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}
$

the first few terms are as follows,

-1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9......

I know I should take into the case of positive and negative terms by themselves. Using the epsilon/delta definition would also help. Thanks

EDIT: Sorry about mixing the variables guys, been a long day
• Oct 27th 2008, 02:43 PM
Plato
Quote:

Originally Posted by Maxmw22
OK, questions is about divergent, conditionally convergent, or abs convergent.
the series is n=1 to infinity of [(-1)^(sqrt(floor of n))]/n

I think that there is a miss-type in that question.
Note that for positive integers, the floor function is $\left\lfloor n \right\rfloor = n \Rightarrow \quad \sqrt {\left\lfloor n \right\rfloor } = \sqrt n$.
From what you wrote “the first few terms are as follows, -1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9...”
I think this is what you mean $\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}$.
• Oct 27th 2008, 02:47 PM
Maxmw22
Quote:

Originally Posted by Plato
I think that there is a miss-type in that question.
Note that for positive integers, the floor function is $\left\lfloor n \right\rfloor = n \Rightarrow \quad \sqrt {\left\lfloor n \right\rfloor } = \sqrt n$.
From what you wrote “the first few terms are as follows, -1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9...”
I think this is what you mean $\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}$.

yes floor of the sqrt, not sqrt of the floor, seems like I am making all kinds of mistakes...haha

but yes your expression is the one I am dealing with.
• Oct 27th 2008, 05:28 PM
Maxmw22
just bumping in hope of some help.
• Oct 28th 2008, 02:38 AM
Laurent
Quote:

Originally Posted by Maxmw22
OK, questions is about divergent, conditionally convergent, or abs convergent.

$
\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}
$

the first few terms are as follows,

-1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9......

It converges. Note $H_N=\sum_{n=1}^N\frac{1}{n}$, and $S_N= \sum\limits_{n = 1}^N {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}$.

The sign changes when $n$ is a square, so that we find, grouping the terms between two squares: $S_{N^2-1}=\sum_{k=1}^{N-1} (-1)^k\left(\frac{1}{k^2}+\frac{1}{k^2+1}+\cdots+\fr ac{1}{(k+1)^2-1}\right) = \sum_{k=1}^{N-1} (-1)^k s_k$ where $s_k=\sum_{n=k^2+1}^{(k+1)^2-1}\frac{1}{n}= H_{(k+1)^2-1}-H_{k^2}$.

The idea is to prove that $\sum_k (-1)^k s_k$ converges. There are different ways to do that:
- using the expansion $H_n=\ln n+\gamma+O_n\left(\frac{1}{n^2}\right)$ to find that $(-1)^n s_n=2\frac{(-1)^n}{n}+O_n\left(\frac{1}{n^2}\right)$ is the sum of the general terms of two convergent series;
- applying directly the alternating series theorem by showing that $(s_n)_{n\geq 1}$ is decreasing and converges to 0. Neither is trivial. To show that $(s_n)_n$ is decreasing, I found (if there's no mistake in my computation) that using comparison with an integral works: show that $s_n$ is less than some integral ( $s_k\leq\int_{k^2-1}^{(k+1)^2-1}\frac{dx}{x}$), and $s_{n+1}$ is greater than another one, which is greater than the previous one (usual comparison with the integral of a decreasing function)... This comparison allows as well to show that $s_n$ converges to 0. [I let you try to fill in the details]

Once you've shown that $\sum_k (-1)^k s_k$ converges, you must deduce that the initial series converges. However, the difference between $S_n$ and $\sum_{k=1}^N (-1)^k s_k$ (where $N^2$ is the greatest square less than $n$) is less than $s_{N+1}$, which converges to 0. There may be mistakes with the indices here, but this is the main idea.
• Oct 28th 2008, 06:22 AM
Maxmw22
Quote:

Originally Posted by Laurent
It converges. Note $H_N=\sum_{n=1}^N\frac{1}{n}$, and $S_N= \sum\limits_{n = 1}^N {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}$.

The sign changes when $n$ is a square, so that we find, grouping the terms between two squares: $S_{N^2-1}=\sum_{k=1}^{N-1} (-1)^k\left(\frac{1}{k^2}+\frac{1}{k^2+1}+\cdots+\fr ac{1}{(k+1)^2-1}\right) = \sum_{k=1}^{N-1} (-1)^k s_k$ where $s_k=\sum_{n=k^2+1}^{(k+1)^2-1}\frac{1}{n}= H_{(k+1)^2-1}-H_{k^2}$.

The idea is to prove that $\sum_k (-1)^k s_k$ converges. There are different ways to do that:
- using the expansion $H_n=\ln n+\gamma+O_n\left(\frac{1}{n^2}\right)$ to find that $(-1)^n s_n=2\frac{(-1)^n}{n}+O_n\left(\frac{1}{n^2}\right)$ is the sum of the general terms of two convergent series;
- applying directly the alternating series theorem by showing that $(s_n)_{n\geq 1}$ is decreasing and converges to 0. Neither is trivial. To show that $(s_n)_n$ is decreasing, I found (if there's no mistake in my computation) that using comparison with an integral works: show that $s_n$ is less than some integral ( $s_k\leq\int_{k^2-1}^{(k+1)^2-1}\frac{dx}{x}$), and $s_{n+1}$ is greater than another one, which is greater than the previous one (usual comparison with the integral of a decreasing function)... This comparison allows as well to show that $s_n$ converges to 0. [I let you try to fill in the details]

Once you've shown that $\sum_k (-1)^k s_k$ converges, you must deduce that the initial series converges. However, the difference between $S_n$ and $\sum_{k=1}^N (-1)^k s_k$ (where $N^2$ is the greatest square less than $n$) is less than $s_{N+1}$, which converges to 0. There may be mistakes with the indices here, but this is the main idea.

thanks, more than likely I will use the alternating series test, as it is the only one defined so far in my book. And of course if it converges it only conditionally converges as it would then resemble the harmonic series. I just need to find a way to rearrange the summands in order to make it converge, I thank you greatly.
• Oct 28th 2008, 03:13 PM
Opalg
Quote:

Originally Posted by Laurent
$s_k=\sum_{n=k^2}^{(k+1)^2-1}\frac{1}{n}$.

The idea is to prove that $\sum_k (-1)^k s_k$ converges.

Another way to prove this (the main step in Laurent's proof) is to show directly that the sequence s_k decreases to 0.

It certainly converges to 0, because $s_k < \sum_{n=k^2}^{(k+1)^2-1}\frac{1}{k^2} = \frac{2k+1}{k^2}\to0$ as $k\to\infty$.

To see that it decreases, we have

$s_k-s_{k+1} = \sum_{n=0}^{2k}\left(\frac1{k^2+n}-\frac1{(k+1)^2+n}\right)- \frac1{(k+2)^2-2} - \frac1{(k+2)^2-1}$

(pairing off the terms in the sum for s_k with those of s_{k+1}, which leaves the two extra terms at the end of s_{k+1} to be subtracted off separately).

This is equal to $\sum_{n=0}^{2k}\frac{2k+1}{(k^2+n)((k+1)^2+n)} - \frac{2(k+2)^2-3}{((k+2)^2-2)((k+2)^2-1)}$, which is greater than $\frac{(2k+1)^2 - 2(k+2)^2+3}{((k+2)^2-2)((k+2)^2-1)} > 0$.
• Oct 29th 2008, 10:13 AM
shawsend
Based on Laurent's analysis above, here's my attempt to prove it converges. I know I can't hit very well but I have a strong arm. Personally I think this problem illustrates a deep and profound property of mathematics: Just how much can we stretch it and still have it converge? There's no clear boundary between convergent and divergent series; a fractal zoo separates them I believe but I digress. Here goes:

\begin{aligned}\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}&= -\left(1+1/2+1/3\right) &=s_1 \\
&+\left(1/4+1/5+1/6+1/7+1/8\right)&=s_2 \\
&-\left(1/9+1/10+\cdots+1/15\right)&=s_3 \\
&+1/16+\cdots+1/24&=s_4 \\
&\vdots \\
&(-1)^k\left(\frac{1}{k^2}+\frac{1}{k^2+1}+\frac{1}{k ^2+2}+\cdots+\frac{1}{(k+1)^2-1}\right)&=s_k
\end{aligned}

so that:

$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}=\sum_{k=1}^{\infty } (-1)^k s_k$

This is now an alternating series and by inspection, it's obvious that:

$\left(\frac{1}{k^2}+\cdots+\frac{1}{(k+1)^2-1}\right)>\left(\frac{1}{(k+1)^2}+\cdots+\frac{1}{ (k+2)^2-1}\right)$

Therefore, $s_{k+1}. Thus by the Alternating series test, if $\lim_{k\to\infty} s_k\to 0$ then the original series converges. But:

$\lim_{k\to \infty}\left\{\int_{k^2}^{(k+1)^2-1}\frac{dx}{x} and thus by the Squeeze Theorem, $s_k\to 0$. Therefore the original series converges.
• Oct 29th 2008, 10:38 AM
Laurent
Quote:

Originally Posted by shawsend
This is now an alternating series and by inspection, it's obvious that:

$\left(\frac{1}{k^2}+\cdots+\frac{1}{(k+1)^2-1}\right)>\left(\frac{1}{(k+1)^2}+\cdots+\frac{1}{ (k+2)^2-1}\right)$

The fact is that this is not obvious, at least for me (there are more terms in the second sum than in the first one). This is the reason for my comparison with integrals (something like $s_k\geq \int(\cdots)=\cdots\geq \cdots = \int(\cdots)\geq s_{k+1}$ with appropriate things in the dots), and this is what OpAlg proves in a different manner (by estimating the difference $s_{k}-s_{k+1}$ directly).

Quote:

Therefore, $s_{k+1}. Thus by the Alternating series test, if $\lim_{k\to\infty} s_k\to 0$ then the original series converges. But:

$\lim_{k\to \infty}\left\{\int_{k^2}^{(k+1)^2-1}\frac{dx}{x} and thus by the Squeeze Theorem, $s_k\to 0$.
For this convergence of $s_k$ to 0, I mentioned the comparison with the integral I had done previously (btw, note that $0\leq s_k$ hence the upper bound is enough), but OpAlg gives a much simpler argument, which consists in bounding each term of the sum defining $s_k$ by the greatest one (i.e. the first one).
• Oct 29th 2008, 10:49 AM
shawsend
Quote:

Originally Posted by Laurent
The fact is that this is not obvious, at least for me (there are more terms in the second sum than in the first one).

Sorry. Dumb mistake for me.

I need to study you and Opalg's analysis more.
• Oct 29th 2008, 12:23 PM
shawsend
Ok I got it now. Sorry for not looking at your post more carefully first Opalg. Honestly if I was in class with either of you I'd just sit in the back row and try my best to keep quiet as I'd have no hope of ever making the highest grade in anything the whole semester. I would of course expect our professor to assign quite challenging problems to both of you; the clowns in the back row, less so.