Infinite Series problem

[ijd5000]

Hi, i'm having trouble with this series in particular evaluating the limit. It appears that it's going to be 1/3, tried l'hospitals rule which just made it a mess.

 

[ijd5000]

sorry the picture cut of my b_n. It's 1/n^{3/2 .}

 

[pka]

Hi, i'm having trouble with this series in particular evaluating the limit. It appears that it's going to be 1/3, tried l'hospitals rule which just made it a mess.

Note that \(\displaystyle \dfrac{{\sqrt {n + 1} }}{{3{n^3} + n + 5}} \le \dfrac{{\sqrt 2 }}{{3\sqrt[3]{{{n^2}}}}}\)

Use the basic comparison test and a p-series.

 

[lookagain]

Hi, i'm having trouble with this series in particular evaluating the limit. It appears that it's going to be 1/3, tried l'hospitals rule which just made it a mess.

If evaluating \(\displaystyle \ \displaystyle\lim_{x \to \infty} \dfrac{a_n}{b_n} \ \ \) is supposed to be part of one of the methods for the summation

of this series, then I'll show some steps for your limit (that you have at the bottom of your paper).


\(\displaystyle \displaystyle\lim_{x\to\infty} \ \dfrac{ \ n^{\frac{3}{2}}(n + 1)^{\frac{1}{2}}}{ \ 3n^2 + n + 5 \ } \ =\)


\(\displaystyle \displaystyle\lim_{x\to\infty} \ \dfrac{ \ (1/n^2)[n^{\frac{3}{2}}(n + 1)^{\frac{1}{2}}]}{ \ (1/n^2)(3n^2 + n + 5) \ } \ = \)


\(\displaystyle \displaystyle\lim_{x\to\infty} \ \dfrac{ \dfrac{n^{\frac{3}{2}}}{n^{\frac{3}{2}}} \cdot \dfrac{(n + 1)^{\frac{1}{2}}}{n^{\frac{1}{2}}}}{ \dfrac{3n^2}{n^2} + \dfrac{n}{n^2} + \dfrac{5}{n^2}} \ = \)


\(\displaystyle \displaystyle\lim_{x\to\infty} \ \dfrac{ \bigg(\dfrac{n}{n} + \dfrac{1}{n}\bigg)^{\frac{1}{2}}}{ 3 + \dfrac{1}{n} + \dfrac{5}{n^2}} \ = \)



\(\displaystyle \displaystyle\lim_{x\to\infty} \ \dfrac{ \bigg(1 + \dfrac{1}{n}\bigg)^{\frac{1}{2}}}{ 3 + \dfrac{1}{n} + \dfrac{5}{n^2}} \ = \)


\(\displaystyle \ \dfrac{ (1 + 0)^{\frac{1}{2}}}{ 3 + 0 + 0} \ = \)


\(\displaystyle \dfrac{1}{3}\)