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$a_n$=$\frac{1}{\sqrt{(n)(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+....+ \frac{1}{\sqrt{(2n-1)(2n)}}$ this sequence can be written as $$a_n=\sum_{k=n}^{2n-1}\frac{1}{\sqrt{k(k+1)}}$$ Now, the $nth$ term of this sequence is a sequence of patial sums of the corresponding series( I think this assumption is wrong because the sum varies over $n$ to $2n-1$ (not starting from $n=1$)) .I can think of two ways to check the convergence

1st We can write $$\frac{1}{\sqrt{n(n+1)}} =\frac{n+1}{\sqrt{n(n+1)}}-\frac{n}{\sqrt{n(n+1)}}$$ and

$$\frac{1}{\sqrt{(n+1)(n+2)}} =\frac{n+2}{\sqrt{(n+1)(n+2)}}-\frac{n+1}{\sqrt{(n+1)(n+2 )}}$$... Similarly $$\frac{1}{\sqrt{(2n-1)(2n)}} =\frac{2n}{\sqrt{(2n-1)(2n)}}-\frac{(2n-1)}{\sqrt{(2n-1)(2n)}}$$ So, $$a_n =\frac{\sqrt{(n+1)}}{\sqrt{n}}-\frac{\sqrt{(n)}}{\sqrt{n+1}}+\frac{\sqrt{(n+2)}}{\sqrt{n+1}}-\frac{\sqrt{(n+1)}}{\sqrt{n+2}}+...\frac{\sqrt{(2n)}}{\sqrt{2n-1}}-\frac{\sqrt{(2n-1)}}{\sqrt{n+n}}$$ When $\lim_{n\to∞}$ $$a_n=(1-1)+(1-1)+.... +(1-1)$$ so $\lim_{n\to∞}$ is zero.

2nd (I have doubt in this) can we use series comparison test here?

Take $$u_n=\frac{1}{\sqrt{k(k+1)}}$$ & $$v_n= 1/n$$ $$\lim_{n\to\inf}=u_n/v_n=1\text{(finite & non zero)}$$

Hence $\sum{u_n}$ and $\sum{v_n}$ behaves alike , since $\sum (1/n)$ is Divergent,so $\sum{u_n}$ is. Since I have already mentioned that the summation starts varies from $n$ (not from 1) , is this reason is okay or can someone provide a better explanation for this ?

Gajjze
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    First approach is not true because adding property of limit operation is true only for finite number of terms, here number of terms also goes to infinity. Simpler example is $1 = n\cdot\frac{1}{n}=\frac{1}{n}+\dots+\frac{1}{n}\to 0$ as every term goes to $0$ – Egor Ivanov Jan 09 '24 at 11:17
  • Egor is right, the first method is incorrect because the number of terms in the series tends to infinity. The second method is also incorrect because the comparison tests start from $k=$ constant, not $k=f(n).$ – Adam Rubinson Jan 09 '24 at 12:50

2 Answers2

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Khostrotash has already showed that $a_n$ is bounded below by $\frac{1}{2}.$

All that is left to show is that $(a_n)$ is a decreasing sequence, and then by the monotone convergence theorem, we will have proved that $\displaystyle\lim_{n\to\infty} a_n$ converges. To show that $(a_n)$ is decreasing means to show that $a_n \geq a_{n+1},\ $ i.e. $a_n - a_{n+1}\geq 0.$ We have:

$$ a_n - a_{n+1} = \sum_{k=n}^{2n-1}\frac{1}{\sqrt{k(k+1)}} - \sum_{k=n+1}^{2n+1}\frac{1}{\sqrt{k(k+1)}}$$

$$ = \left( \frac{1}{\sqrt{n(n+1)}} + \require{cancel}\cancel{\frac{1}{\sqrt{(n+1)(n+2)}} } + \ldots + \require{cancel}\cancel{\frac{1}{\sqrt{(2n-1)(2n)}} } \right)$$

$$ - \left( \require{cancel}\cancel{\frac{1}{\sqrt{(n+1)(n+2)}} } + \ldots + \require{cancel}\cancel{ \frac{1}{\sqrt{(2n-1)(2n)}} } + \frac{1}{\sqrt{(2n)(2n+1)}} + \frac{1}{\sqrt{(2n+1)(2n+2)}} \right) $$

$$ = \frac{1}{\sqrt{n(n+1)}} - \left( \frac{1}{\sqrt{(2n)(2n+1)}} + \frac{1}{\sqrt{(2n+1)(2n+2)}} \right). $$

We want to show this is $\geq 0,$ i.e.

$$ \frac{1}{\sqrt{n(n+1)}} \geq \frac{1}{\sqrt{(2n)(2n+1)}} + \frac{1}{\sqrt{(2n+1)(2n+2)}} \tag{1}\label{eq1}. $$

Squaring the RHS gives:

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

$$ + 2\left(\frac{1}{2n+1}\right) \sqrt{ \left(\frac{1}{2n}\right) \left( \frac{1}{2n+2}\right) } $$

$$ \leq \frac{1}{(2n)(2n+1)} + \frac{1}{(2n+1)(2n+2)} + \require{cancel}\cancel{2}\left(\frac{1}{2n+1}\right) \frac{ \left( \frac{1}{2n} + \frac{1}{2n+2} \right)}{\require{cancel}\cancel{2}}\quad \text{ (AM-GM)} $$

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

$$ = 2\left( \frac{1}{(2n)} - \require{cancel}\cancel{\frac{1}{(2n+1)} } + \require{cancel}\cancel{ \frac{1}{(2n+1)} } - \frac{1}{(2n+2)} \right) = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}. $$

Alternative approach, which I'll write without rigour for brevity's sake, but it should still convince those familiar with results of harmonic series. For those unfamiliar, then see this result for example.

From

$$ \frac{1}{\sqrt{(k+1)(k+1)}} \leq \frac{1}{\sqrt{k(k+1)}} \leq \frac{1}{\sqrt{k\cdot k}}, $$

it is not too hard to use the result in the link above and the Squeeze theorem and show that $\displaystyle\lim_{n\to\infty} a_n = \ln 2.$

Adam Rubinson
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  • I was trying to do so but getting stuck at the last step that when $\lim_{n \to inf}$ both sides of the expression approaches $0$. But now clear. Thank you. – Gajjze Jan 09 '24 at 15:46
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Note that $$a_n=\frac{1}{\sqrt{(n)(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+....+ \frac{1}{\sqrt{(2n-1)(2n)}} \\ \ge \frac{1}{\sqrt{(2n-1)(2n)}}+\frac{1}{\sqrt{(2n-1)(2n)}}+....+ \frac{1}{\sqrt{(2n-1)(2n)}}\\=n\times\frac{1}{\sqrt{(2n-1)(2n)}} \\ \ge n\times\frac{1}{\sqrt{(2n)(2n)}}\\=\frac{n}{\sqrt{(2n)^2}}=\frac 12$$

Khosrotash
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