Question

Calculate the following limit without using

L'Hôpital's rule

`\rm\underset{n \to \infty}{\lim} \frac{ \sqrt[3]{ \left( 1 + \frac{1}{ \sqrt[5]{2} } + \frac{1}{\sqrt[5]{3} } + \dots \frac{1}{\sqrt[5]{n}}\right)^(2) } }{\sqrt[5]{ \left( 1 + \frac{1}{ \sqrt[3]{2} } + \frac{1}{\sqrt[3]{3} } + \dots \frac{1}{\sqrt[3]{n}}\right)^(4) }}`
​

Answer 1

For brevity, let

`S_i = \displaystyle \sum_(k=1)^n \frac1{\sqrt[i]{k}}`

Rewrite the limit as

`\displaystyle \lim_(n\to\infty) \frac{{S_5}^(\frac23)}{{S_3}^(\frac45)} = \lim_(n\to\infty) \frac{n^(\frac23) \left(\frac{S_5}n\right)^(\frac23)}{n^(\frac45) \left(\frac{S_3}n\right)^(\frac45)} = \lim_(n\to\infty) \frac1{n^{\frac2{15}}} * (\left(\lim\limits_(n\to\infty)\left(\frac1n S_5\right)\right)^(\frac23))/(\left(\lim\limits_(n\to\infty)\left(\frac1n S_3\right)\right)^(\frac45))`

The two limits involving S are equivalent to definite integrals,

`\displaystyle \lim_(n\to\infty) \frac1n S_i = \lim_(n\to\infty) \frac1n \sum_(k=1)^n \frac1{\sqrt[i]{\frac kn}} = \int_0^1 \frac1{\sqrt[i]{x}} \, dx = \frac i{i-1}`

Then the end result is

`\displaystyle 0 * (\left(\frac54\right)^(\frac23))/(\left(\frac32\right)^(\frac45)) = \boxed{0}`