To show: Sequence $\langle u_n\rangle$ is convergent and converges to zero,where $u_n=\frac1{1n} + \frac1{2(n-1)} +\frac1{3(n-2)}+\dots+\frac1{n1}$

Question

To show: Sequence $\langle u_n\rangle$ is convergent and converges to zero,where $u_n=\frac{1}{1n} + \frac{1}{2(n-1)} +\frac{1}{3(n-2)}+ \dots +\frac{1}{n1}$

Using Rabee's test I have shown that the sequence $\langle u_n\rangle$ is convergent.
Now to find the limit: $\frac{1}{nn} + \frac{1}{nn} +\frac{1}{nn}+ \dots +\frac{1}{nn}\le \frac{1}{1n} + \frac{1}{2(n-1)} +\frac{1}{3(n-2)}+ \dots +\frac{1}{n1}$
$\implies\frac{1}{n^2} + \frac{1}{n^2} +\frac{1}{n^2}+ \dots +\frac{1}{n^2}\le \frac{1}{1n} + \frac{1}{2(n-1)} +\frac{1}{3(n-2)}+ \dots +\frac{1}{n1}$
$\implies\frac{1}{n}\le \frac{1}{1n} + \frac{1}{2(n-1)} +\frac{1}{3(n-2)}+ \dots +\frac{1}{n1}$
I am unable to find an upper bound,which converges to zero.
Kindly help me. Also,if there are different methods then kindly share.

$$\sum\limits_{k=0}^{n-1} \frac{1}{(n-k)(k+1)}\sum\limits_{k=0}^{n-1} \frac{-1}{((k+1)-(n+1))(k+1)}=\frac{1}{n+1}\sum\limits_{k=0}^{n-1} \frac{1}{k+1}-\frac{1}{k-n}$$ — pie, Oct 01 '24 at 09:57
@SandeepTiwari the first sum in the right hand side is about $\ln(n)$ use l,hobital and the second one is also about $\ln(n)$ take the absolute value — pie, Oct 02 '24 at 11:43
@pie Sorry sir after so many hints,I am still not getting it.What is l hobital? — San Ti, Oct 02 '24 at 11:53
@pie Kindly post an answer,it will be helpful.Thank you sir. — San Ti, Oct 02 '24 at 11:57
@pie The same hint is given in other post too but persons like me will never reach the solution. — San Ti, Oct 02 '24 at 11:58
@TonyK "Thank you for your advice, but by writing it this way, the next term of the series becomes easy to figure out." — San Ti, Oct 08 '24 at 03:27

score 2 · Accepted Answer · answered Oct 02 '24 at 12:30

2

$$\frac{1}{x(x-y)}=\frac{1}{y}\left(\frac{1}{x-y}-\frac{1}{x}\right)$$

$$u_n=\sum \limits_{k=0}^ {n-1}\frac{1}{(k+1)(n-k)} =\sum \limits_{k=0}^ {n-1}\frac{1}{(k+1)((n+1)-(k+1))}$$

$$=\frac{1}{n+1}\left(\sum \limits_{k=0}^ {n-1}\frac{1}{(k+1)}+\sum \limits_{k=0}^ {n-1}\frac{1}{(n-k)}\right)$$

$$= \frac{1}{n+1}\left(\sum \limits_{k=0}^ {n-1}\frac{1}{(k+1)}+\sum \limits_{m=0}^ {n-1}\frac{1}{(1+m) )}\right)$$

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

$$\lim\limits_{n \to \infty } u_n =\lim\limits_{n \to \infty } \frac{2}{n+1}\sum_{k=1}^n \frac{1}{k} =\lim\limits_{n \to \infty } \frac{2\ln(n)}{n} $$ By this or This

answered Oct 02 '24 at 12:30

pie

8,483

Sir,how you got last equality? – San Ti Oct 02 '24 at 23:47
@SandeepTiwari https://en.wikipedia.org/wiki/Harmonic_number, $$\lim\limits_{n \to \infty }\sum_{k=1}^n\left(\frac 1 k\right)-\ln(n) =\gamma$$
where $\gamma$.
– pie Oct 03 '24 at 15:03
@SandeepTiwari which one do you mean? – pie Oct 03 '24 at 15:15
I am getting this, $$\lim\limits_{n \to \infty } \frac{2}{n+1}\sum_{k=1}^n \frac{1}{k} =\lim\limits_{n \to \infty } \frac{2(\gamma+\ln(n)}{n+1} $$ – San Ti Oct 04 '24 at 14:10
but it solved the problem.Thank you so much. – San Ti Oct 04 '24 at 14:10

score 1 · Answer 2 · answered Oct 07 '24 at 16:26

$$\frac{1}{n}\le u_n=\frac{2}{n+1}\sum_{k=1}^n \frac{1}{k}\le\frac{2}{n}\sum_{k=1}^n \frac{1}{k}$$Now,let $a_n=\frac{1}{n}$ and $b_n=\frac{2}{n}\sum_{k=1}^n \frac{1}{k}$.
Then,$\langle a_n\rangle\rightarrow 0$ and by Cauchy's first theorem on limits $\langle b_n\rangle\rightarrow 0$.
Thus,by Squeeze theorem $\langle u_n\rangle\rightarrow 0$

To show: Sequence $\langle u_n\rangle$ is convergent and converges to zero,where $u_n=\frac1{1n} + \frac1{2(n-1)} +\frac1{3(n-2)}+\dots+\frac1{n1}$

2 Answers2

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