convergence of series

Anil

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determine whether or not the series n starts 1 end go to infinity, (In(n)+((-1)^n )n^(1/2))/ n*n^(1/2) is convergent? I couldnt do it
 
determine whether or not the series n starts 1 end go to infinity, (In(n)+((-1)^n )n^(1/2))/ n*n^(1/2) is convergent? I couldnt do it

It is not In, it is ln, lower case L. You may first want to use the fact that sqrt(n) > ln(n) for all n.
 
Hello, Anil

Is there a typo?
As given, the problem looks quite silly . . .


\(\displaystyle \displaystyle \sum^{\infty}_{n=1} \left[\ln(n)+(-1)^n\frac{n^{\frac{1}{2}}}{n\cdot n^{\frac{1}{2}}}\right]\)

Why didn't they cancel the \(\displaystyle \frac{n^{\frac{1}{2}}}{n^{\frac{1}{2}}}\) ?
 
determine whether or not the series n starts 1 end go to infinity, (In(n)+((-1)^n )n^(1/2))/ n*n^(1/2) is convergent? I couldnt do it
Or might it be

\(\displaystyle \displaystyle \sum^{\infty}_{n=1} \left[\frac{ \ln(n)+(-1)^n n^{\frac{1}{2}}}{n\cdot n^{\frac{1}{2}}}\right]\) ??

[that is what I get when I match up the parentheses in the inline statement]
 
determine whether or not the series n starts 1 end go to infinity,

(In(n)+((-1)^n )n^(1/2))/ n*n^(1/2) **

In any event,the expression is typed incorrectly,
because it is missing a last pair of grouping symbols.



is convergent? I couldnt do it


* * (ln(n) + ((-1)^n)n^(1/2))/(n*n^(1/2))
 
It You may first want to use the fact that sqrt(n) > ln(n) for all (positive integers) n.

But, \(\displaystyle \dfrac{3}{4}(\sqrt{n}) \ > \ ln(n),\) too, \ \ for all positive integers n.


So, it must be established that \(\displaystyle ln(n) \ = \ n^{\frac{1}{2} - \epsilon}, \ \ where \ \ 0 < \epsilon < 1,\)

and where n belongs to the set of all positive integers.


It is not sufficient that \(\displaystyle \sqrt{n} \ > \ ln(n), \) for all positive integers n,

to have \(\displaystyle \ \displaystyle\sum_{n = 1}^{\infty}\dfrac{ln(n)}{n\cdot n^{\frac{1}{2}}} \ \) be convergent.
 
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