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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

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    $\begingroup$ Are those the only two terms given? It is not clear what $...$ means. $\endgroup$
    – N. S.
    Jan 25, 2020 at 4:13
  • $\begingroup$ @N.S. Yes they are. It looks exactly like this on the exam. I can post a picture if it helps. $\endgroup$
    – tau20
    Jan 25, 2020 at 4:15
  • $\begingroup$ Can you give this problem in summation form? $\endgroup$
    – imranfat
    Jan 25, 2020 at 4:15
  • $\begingroup$ @imranfat The problem looks exactly like this on the exam. Therefore, I am unsure of the summation form. $\endgroup$
    – tau20
    Jan 25, 2020 at 4:18

6 Answers 6

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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,

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

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  • $\begingroup$ I see, the squeeze theorem is to be deployed. I fail to understand how did you get the smallest and the largest expressions in the ordering. Could you elaborate ? Thanks for a nice answer $\endgroup$
    – tau20
    Jan 25, 2020 at 5:20
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    $\begingroup$ @GrigoriPerelman : I just saw i missed a factor on the LHS. The estimation comes as follows: Each summand is of the form $\frac{1}{\sqrt[4]{1+\frac{n+k+1}{n^4}}}$. For each summand we have $$\frac{1}{\sqrt[4]{1+\frac{5}{n^3}}}=\frac{1}{\sqrt[4]{1+\frac{5n}{n^4}}}< \frac{1}{\sqrt[4]{1+\frac{n+k+1}{n^4}}} < \frac{1}{\sqrt[4]{1+\frac{n}{n^4}}} = \frac{1}{\sqrt[4]{1+\frac{1}{n^3}}}$$ $\endgroup$ Jan 25, 2020 at 5:30
  • $\begingroup$ I see, but I miserably fail to understand why do you have 4n-2 in the numerators on the LHS and RHS in the parent answer ? $\endgroup$
    – tau20
    Jan 25, 2020 at 5:39
  • $\begingroup$ Because there are $4n-2$ summands altogether from $k=1$ to $k=4n-2$. $\endgroup$ Jan 25, 2020 at 5:40
  • $\begingroup$ Let me ask you one more thing. Did you get the n^4+n+1+k = n^4+5n-1-(4n-2)+k this way ? Is this an usual rule to determine that term in the series ? $\endgroup$
    – tau20
    Jan 25, 2020 at 5:59
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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 $$\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}} \leq \\ \leq \frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+3}}+\frac{1}{\sqrt[4]{{n^4}+n+4}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}} \leq \\ \leq \frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+n+2}}$$

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  • $\begingroup$ Thanks for your contribution. I guess the squeeze is the best option. I am going to try to finish it. P.S. Could you post in the edit the next step, how would it look like the limit evaluation of the smallest(or largest) sum in the ordering? $\endgroup$
    – tau20
    Jan 25, 2020 at 4:56
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$$L=\lim_{n \rightarrow \infty} \sum_{k=1}^{4n-1} \frac{1}{(n^4+n+k)^{1/4}}$$ $$L=\lim_{n \rightarrow \infty} \sum_{k=1}^{4n-1}\frac{1}{n} \frac{1}{(1+\frac{1}{n^3}+\frac{k}{n^4})^{1/4}}= \int_{0}^{4} dx=4.$$ Here $1/n^3$, $k/n^3$ can be neglected. Also term $1/(n^4+n+1)^{1/4} \rightarrow 0$ as $n \rightarrow \infty$.

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  • $\begingroup$ Wow, you beat me to it! $\endgroup$
    – Kagaratsch
    Jan 25, 2020 at 4:42
  • $\begingroup$ Thanks a lot for you contribution, but I forgot to emphasize that the limit is supposed to be solved only with the tools prior to derivatives and integrals. I made an edit in the post so that is clear now. $\endgroup$
    – tau20
    Jan 25, 2020 at 4:43
  • $\begingroup$ Not conclusive. You should explain WHY those terms can be neglected. Hint: Sandwich. $\endgroup$ Jan 25, 2020 at 4:47
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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 $$0<n^4< n^4+j\le$$ $$\le n^4 +5n-2 <$$ $$< n^4+4n+6n+4n+1\le$$ $$\le n^4+4n^3+6n^2+4n+1=(n+1)^4.$$

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  • $\begingroup$ It is not always clear what "$+...+$" means, but the $\sum$ notation, when used properly, is not ambiguous. $\endgroup$ Jan 25, 2020 at 4:59
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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.

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  • $\begingroup$ Thanks a lot for you contribution, but I forgot to emphasize that the limit is supposed to be solved only with the tools prior to derivatives and integrals. I made an edit in the post so that is clear now. $\endgroup$
    – tau20
    Jan 25, 2020 at 4:43
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    $\begingroup$ @GrigoriPerelman see my updated answer. $\endgroup$
    – Kagaratsch
    Jan 25, 2020 at 4:53
  • $\begingroup$ "the coefficient of n in the denominator goes from 1⋅n to 5⋅n, which means that roghly 4n terms are being summed". Could you elaborate on how did you conclude that roughly 4n terms are being summed, i.e. how did you find out that uper boundary of the sigma ? Thanks for the answer, btw I am not familiar with the O-sign notation, so I guess I fail to understand your argument fully. $\endgroup$
    – tau20
    Jan 25, 2020 at 5:00
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    $\begingroup$ @GrigoriPerelman see updated answer, the big O notation is not really necessary here, so I removed it. $\endgroup$
    – Kagaratsch
    Jan 25, 2020 at 5:13
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    $\begingroup$ @GrigoriPerelman I rewrote it a little bit. Basically, the problem is reduced from taking a sum, to showing the limit on each summand. Taking into account the possible values of $i$ you can fill in the details. $\endgroup$
    – Kagaratsch
    Jan 25, 2020 at 5:30
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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 $1<k\le 4n-1$

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