# calculating 2 sums of series

So I have these two series given.

1:

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

And I have to show that this sum is $\leq$ 1.

2:

$\displaystyle\sum_{n=1}^{\infty}\frac{n-\sqrt{n^2-1}}{\sqrt{n^2+n}}$

And I have to calculate this sum.

So with first series I tried with partial sums but it just didn't go well. Should I use partial fractions? Then perhaps I can go through with it. But with the second one I thought of just going partial sums, but I got stuck, because nothing seems to be reducing or I made a calclulating error.

So any help would be appreciated.

• The second sum doesn't seem to converge, as it asymptotically behaves like $1/n$. – glS Jun 29 '15 at 13:05
• I made a mistake while writing in Latex :/ i made an edit. sorry my bad. – MathIsTheWayOfLife Jun 29 '15 at 13:26
• That completely changes the problem... – Jack D'Aurizio Jun 29 '15 at 13:30
• FYI, the plural of "series" is "series". – woz Jun 29 '15 at 13:44
• Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Jun 29 '15 at 16:05

HINT:

For the first one,

$$\sin(2n)\le1$$

and consequently, $$\dfrac{\sin2n}{n(n+1)}\le\dfrac1{n(n+1)}$$

Now $$\dfrac1{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\cdots$$

• can you give me the next step maybe? i really don't know what to do next. – MathIsTheWayOfLife Jun 29 '15 at 12:50
• @user246608, Set $n=1,2,3,\cdots,m$ to find the partial sum – lab bhattacharjee Jun 29 '15 at 13:29
• I already solved it, a bit differently, look in another answer. It is easier and faster. And we did it normally both ways i think. So thankyou. – MathIsTheWayOfLife Jun 29 '15 at 13:33
• @user246608, Welcome! Which is meant here when you say "easier and faster" – lab bhattacharjee Jun 29 '15 at 13:39
• well mather of "easier" is subjective so i shouldn't really be saying it. Faster was i think because of the way it was calculated. But it was a simmilar concept look at ChickenP answer, it's the way we solved it during class. – MathIsTheWayOfLife Jun 29 '15 at 13:43
1. $$\frac{n-\sqrt{n^2 -1} }{\sqrt{n(n+1)}} =\frac{n}{\sqrt{n(n+1)} }-\frac{\sqrt{n^2-1}}{\sqrt{n(n+1)}} =\sqrt{\frac{n}{n+1}} -\sqrt{\frac{n-1}{n}}$$ hence $$\sum_n \frac{n-\sqrt{n^2 -1} }{\sqrt{n(n+1)}} =\sum_n \left(\sqrt{\frac{n}{n+1}} -\sqrt{\frac{n-1}{n}} \right) =\lim_{n\to\infty }\sqrt{\frac{n}{n+1}} =1$$ thus $$\sum_n \frac{n-\sqrt{n^2 -1} -1}{\sqrt{n(n+1)}} = \sum_n\frac{n-\sqrt{n^2 -1} }{\sqrt{n(n+1)}} -\sum_n \frac{1}{\sqrt{n(n+1)}} =1-\sum_n \frac{1}{\sqrt{n(n+1)}}$$ so the problem is equivalent with problem to find the following sum $$\sum_n \frac{1}{\sqrt{n(n+1)}} .$$
• is that at the end in first row really legible to do? – MathIsTheWayOfLife Jun 29 '15 at 13:00
• Definitely. $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ for non-negative $a, b$. – Zain Patel Jun 29 '15 at 13:02
• The last series is trivially non-converging. – Jack D'Aurizio Jun 29 '15 at 13:05
• @Jack D'Aurizio You're right. – MotylaNogaTomkaMazura Jun 29 '15 at 13:09
• @Jack D'Aurizio I did a mistake while typing in Latex, it's edited now. It should work now. – MathIsTheWayOfLife Jun 29 '15 at 13:29

The first series is trivially absolutely convergent and bounded by the Mengoli series. Moreover: $$\begin{eqnarray*}\sum_{n=1}^{+\infty}\frac{\sin(2n)}{n(n+1)}&=&\text{Im}\sum_{n\geq 1}\left(\frac{e^{2in}}{n}-\frac{e^{2in}}{n+1}\right)\\&=&\color{blue}{(\pi-2)\sin^2(1)-\sin(2)\log(2\sin 1)}\\[0.2cm]&=&0.3350026\ldots\end{eqnarray*}$$ by recognizing a well-known Fourier series. The second series is telescopic: $$\sum_{n\geq 1}\frac{n-\sqrt{n^2-1}}{\sqrt{n}\sqrt{n+1}}=\sum_{n\geq 1}\left(\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n-1}{n}}\right)=\lim_{n\to +\infty}\sqrt{\frac{n}{n+1}}=\color{blue}{1}.$$

• I should've said in beforehand that I am in 1st semester of Calculus, we didn't learned Menoli or Fourier series yet, thus we can't use it. – MathIsTheWayOfLife Jun 29 '15 at 13:23
• @user246608: despite esotic names, the Mengoli series is just the very first telescopic series one computes, namely $\sum_{n\geq 1}\frac{1}{n(n+1)}$. I just used Fourier series to compute the series in closed form (giving a stronger bound), but that wasn't requested to you. – Jack D'Aurizio Jun 29 '15 at 13:27
• okay, i get it now. – MathIsTheWayOfLife Jun 29 '15 at 13:28

For the first summation, we know that $\sin 2n \leq 1$ so $$\frac{\sin2n}{n(n+1)}\le\frac1{n(n+1)}$$ $$\implies \displaystyle\sum_{n=1}^{\infty}\frac{\sin{(2n)}}{n(n+1)} \leq \sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right)$$ But we know that $$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)} = \sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right) = 1$$ So $$\displaystyle\sum_{n=1}^{\infty}\frac{\sin{(2n)}}{n(n+1)} \leq 1$$

For the second : Use $$\frac{n-\sqrt{n^2-1}}{\sqrt{n^2+n}}=\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n-1}{n}}.$$

The following is for the previous version of the second question.

For $n\gt 1$,

\begin{align}\frac{n-1-\sqrt{n^2-1}}{\sqrt{n^2+n}}&=\frac{n-1-\sqrt{n^2-1}}{\sqrt{n^2+n}}\cdot \frac{n-1+\sqrt{n^2-1}}{n-1+\sqrt{n^2-1}}\\&=\frac{-2(n-1)}{(n-1+\sqrt{n^2-1})\sqrt{n^2+n}}\\&\le -2\cdot\frac{1}{\left(1+\frac{\sqrt{n^2-1}}{n-1}\right)\sqrt{n^2+2n+1}}\\&= -2\cdot\frac{1}{\left(1+\frac{\sqrt{n^2-1}}{n-1}\right)(n+1)}\\&\le -2\cdot\frac{1}{(1+\sqrt 3)(n+1)}\end{align}

So, we have $$\sum_{n=1}^{\infty}\frac{n-1-\sqrt{n^2-1}}{\sqrt{n^2+n}}=-\infty.$$

• i made a mistake whiel writing in latex, it's edited now. Please edit answer. Sorry – MathIsTheWayOfLife Jun 29 '15 at 13:27

For the second question

Let $t_n$ be the n-th term in the summation, and S the overall sum. We have

$t_n = \frac{n - \sqrt{n^2 - 1}}{\sqrt{n^2 + n}} \times \frac{n + \sqrt{n^2 - 1}}{n + \sqrt{n^2 - 1}} = \frac{n^2 - (n^2 - 1)}{\sqrt{n^2 + n} \cdot (n + \sqrt{n^2 - 1})} = \frac{1}{\sqrt{n^2 + n} \cdot (n + \sqrt{n^2 - 1})}$

Now for $n \ge 1$, $n + \sqrt{n^2 - 1} \ge n$ and $\sqrt{n^2 + n} > n$, so

$0 < t_n < \frac{1}{n \times n} = \frac{1}{n^2}$

This shows that $0 < S < \frac{\pi^2}{6}$ and that S is convergent (since $\lim\limits_{n \rightarrow \infty}{t_n} = 0$).

We can rewrite the summation, and then rewrite as a limit

$S = \sum\limits_{n=1}^{\infty}{(\sqrt{\frac{n}{n+1}} - \sqrt{\frac{n-1}{n}}}) = \lim\limits_{k \rightarrow \infty}{(\sum\limits_{n=1}^{k}{\sqrt{\frac{n}{n+1}}} - \sum\limits_{n=1}^{k}{\sqrt{\frac{n-1}{n}}})}$

where

$\sum\limits_{n=1}^{k}{\sqrt{\frac{n-1}{n}}} = 0 + \sum\limits_{n=2}^{k}{\sqrt{\frac{n-1}{n}}} = \sum\limits_{n=1}^{k-1}{\sqrt{\frac{n}{n+1}}}$

so

$S = \lim\limits_{k \rightarrow \infty}{(\sum\limits_{n=1}^{k}{\sqrt{\frac{n}{n+1}}} - \sum\limits_{n=1}^{k-1}{\sqrt{\frac{n}{n+1}}})} = \lim\limits_{k \rightarrow \infty}{\sqrt{\frac{k}{k+1}}} = 1$