Convergence and divergence of a series

[smart_worker]

B]1. Homework Statement [/B]
Find whether the series is convergent or divergent

Homework Equations

The Attempt at a Solution

By ratio test I have,

I would apply L'Hôpital's rule to find the value of limit but before that how do i simplify the expression? It has fractional part both in the numerator as well as in the denominator.

 

[Ray Vickson]

B]1. Homework Statement [/B]
Find whether the series is convergent or divergent

Homework Equations

View attachment 74254

The Attempt at a Solution

By ratio test I have,
View attachment 74255

I would apply L'Hôpital's rule to find the value of limit but before that how do i simplify the expression? It has fractional part both in the numerator as well as in the denominator.

Just use elementary high-school algebra. Alternatively, look more carefully at the problem before even starting. Maybe the ratio test won't work; there are times when it doesn't.

 

[RUber]

Consider breaking it into two sums...the sum of convergent series is convergent, however if one diverges, the sum of the two diverges (generally).

 

[statdad]

If you want to simplify the large fraction in
[tex]
\lim_{n\to \infty} \left(\dfrac{\left(\dfrac{(n+1)^2}{2^{n+1}} + \dfrac{1}{(n+1)^2}\right)}{\dfrac{n^2}{2^n} + \dfrac 1 {n^2}}\right)
[/tex]

treat it the way you would a complex fraction. As has been stated above, however, I'm not sure this approach will generate a positive result.

Think about the idea that if both [itex] \sum_{i=1}^\infty a_n[/itex] and [itex] \sum_{i=1}^\infty b_n [/itex] are absolutely convergent then
[itex] \sum_{i=1}^\infty \left(a_n + b_n \right) [/itex] is absolutely convergent.

 

[LCKurtz]

Consider breaking it into two sums...the sum of convergent series is convergent, however if one diverges, the sum of the two diverges (generally).

Not sure what you mean by "generally" other than perhaps it means "sometimes" because you know it's false in general.