Limit of a sequence using Squeezing Principle: $\sum_{k=1}^n \frac{n}{n^3+k} = \frac{n}{n^3+1}+...+\frac{n}{n^3+n}$.

Question

I have a question involving the following; $$\sum_{k=1}^n \frac{n}{n^3+k} = \frac{n}{n^3+1}+...+\frac{n}{n^3+n}$$ In this problem, I am required to find the limit of the sequence by finding both upper and lower bounds and using the Squeezing Principle.

I would appreciate if anyone would be able to answer this and tell me how to obtain the upper and lower bounds. Thanks

Answer 1

The best idea, while using the squeeze principl for this question, is seeing that your terms of the sum have the same numerator. Therefore, the term with lowest denominator is the biggest term, while the term with largest denominator is the smallest term.

Therefore, it makes sense to try (it mayn't work out) bounding every term by the smallest and largest term to get upper and lower bounds , respectively.

That is, $$ \frac{n}{n^3 + 1} + \ldots + \frac{n}{n^3 + n} \leq n \times \frac{n}{n^3 + 1} = \frac{n^2}{n^3 + 1} \\ \frac{n}{n^3 + 1} + \ldots + \frac{n}{n^3+n} \geq n \times \frac{n}{n^3 + n} = \frac{n^2}{n^3 + n} $$

Thus, the given sequence can be sandwiched between $\frac{n^2}{n^3+1}$ and $\frac{n^2}{n^3+n}$. Do these have limits as $n \to \infty$?

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They do : dividing by $n^2$ on top and bottom gives $\frac{1}{n + \frac 1{n^2}}$ and $\frac{1}{1 + \frac 1n}$, both of which have the limit zero as $n \to \infty$ by the limit theorem for division.

Consequently, the desired limit is also zero by the squeeze theorem.

Answer 2

$\frac n {n^{3}+k} \leq \frac n {n^{3}}=\frac 1 {n^{2}}$ and $\frac n {n^{3}+k} \geq \frac n {n^{3}+n^{3}}=\frac 1 {2n^{2}}$. Since $\sum_{k=1}^{n} \frac 1 {n^{2}}=\frac 1 n$ and $\sum_{k=1}^{n} \frac 1 {2n^{2}}=\frac 2 n$ both have limit $0$, the given sequence also has limit $0$.