Evaluate $\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$

Question

I am new to analysis and I have no clue how to solve this limit. This is an exam problem from my analysis 1 course, there are one or two similar ones on the exam.

$$\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$$

The only thing I tryed was this silly idea to rewrite it as one single fraction and apply Stolz-Cesaro theorem, but it got way too messy so I doubt that is the way.

I can't find explanations generally on these limits of sequences of the type $\frac{1}{f(x_n)}+\cdots+\frac{1}{f(x_{n+k})}$ (I hope this is a good representation). Should series be involved in solving these kinds of limits ?

EDIT: The limit is supposed to be solved only with the knowledge prior to derivatives and integrals.

Thanks in advance

Answer 1

We have

$$\sum_{k=1}^{4n-2}\frac{1}{\sqrt[4]{n^4+n+k+1}} = \frac 1n \sum_{k=1}^{4n-2}\frac{1}{\sqrt[4]{1+\frac{n+k+1}{n^4}}}$$

Hence,

$$\underbrace{\frac{4n-2}{n\sqrt[4]{1+\frac{5}{n^3}}}}_{\stackrel{n\to\infty}{\longrightarrow}4} < \frac 1n \sum_{k=1}^{4n-2}\frac{1}{\sqrt[4]{1+\frac{n+k+1}{n^4}}} < \underbrace{\frac{4n-2}{n\sqrt[4]{1+\frac 1{n^3}}}}_{\stackrel{n\to\infty}{\longrightarrow}4}$$

Answer 2

Hint Assuming that the sum means $$\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+3}}+\frac{1}{\sqrt[4]{{n^4}+n+4}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$$

Use that $$\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}} \leq \\ \leq \frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+3}}+\frac{1}{\sqrt[4]{{n^4}+n+4}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}} \leq \\ \leq \frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+n+2}}$$

Answer 3

$$L=\lim_{n \rightarrow \infty} \sum_{k=1}^{4n-1} \frac{1}{(n^4+n+k)^{1/4}}$$ $$L=\lim_{n \rightarrow \infty} \sum_{k=1}^{4n-1}\frac{1}{n} \frac{1}{(1+\frac{1}{n^3}+\frac{k}{n^4})^{1/4}}= \int_{0}^{4} dx=4.$$ Here $1/n^3$, $k/n^3$ can be neglected. Also term $1/(n^4+n+1)^{1/4} \rightarrow 0$ as $n \rightarrow \infty$.

Answer 4

If $n\ge 1$ then $$4-\frac {2}{n}=\frac {4n-2}{n}=\sum_{j=n+1}^{5n-2}\frac {1}{n}>$$ $$>\sum_{j=n+1}^{5n-2}\frac {1}{(n^4+j)^{1/4}}>$$ $$>\sum_{j=n+1}^{5n-2}\frac {1}{n+1}=$$ $$=\frac {4n-2}{n+1}=4-\frac {6}{n+1}$$ because if $n\ge 1$ and $n+1\le j\le 5n-2$ then $$0<n^4< n^4+j\le$$ $$\le n^4 +5n-2 <$$ $$< n^4+4n+6n+4n+1\le$$ $$\le n^4+4n^3+6n^2+4n+1=(n+1)^4.$$

Answer 5

To figure out how many terms are in the sum $$\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+3}}+\frac{1}{\sqrt[4]{{n^4}+n+4}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$$

take the $n$ dependence of the last term, and subtract from it the $n$ dependence of the first term. Since we should also count the first term itself we add $1$ to this difference and get the number of terms:

$$({n^4}+5n-1)-({n^4}+n+2)+1=4n-2$$

This leads to the sum

$$\lim_{n\to \infty}\sum_{i=1}^{4n-2}\frac{1}{\sqrt[4]{{n^4}+n+1+i}}$$

Note that in the limit $n\to\infty$ we have for all $i$:

$$\frac{n}{\sqrt[4]{{n^4}+n+1+i}}\to 1$$

This means that in the limit, each summand can be equivalently replaced by $1/n$, and summing this leading contribution $4n-2$ times, we get

$$\frac{4n-2}{n}\to 4$$

in the strict limit.

Answer 6

Also note that $\frac{1}{n+1}<\frac{1}{n^4+n+k} <\frac{1}{n}$ so $$ \sum_{k=1}^{4n-1} \frac{1}{n+1}<S_n=\sum_{k=1}^{4n-1} \frac{1}{(n^4+n+k)^{1/4}} <\sum_{k=1}^{4n-1} \frac{1}{n}.$$ So, we note that $$ \lim_{n \rightarrow \infty} \frac{4n-1}{n+1}=4= \lim_{n \rightarrow \infty} \frac{4n-1}{n}= \lim_{n \rightarrow \infty} S_n$$

When $n \rightarrow \infty$ the first term of the sequence $1/(n^4+n+2)^{1/4} \rightarrow 0$ it can well be neglected.

Note that $$\frac{1}{(n^4+n+k)^{1/4}} > \frac{1}{n+1}$$ can be checked to be true for all $1<k\le 4n-1$