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I am approaching the question from an inequality perspective. In other words, I just want to see if the equation has an upper bound or lower bound.

After expanding the equation using binomial expansion, I get the term that $$\left(1 + \frac{1}{\sqrt{n}}\right)^n \leq 1 + n^{1/2} + \frac{n}{2} + \frac{n^{3/2}}{2^{2}}$$

For $n^{1/2} + \frac{n}{2} + ...$ $$a = \sqrt{n}, r = \frac{\sqrt{n}}{2}$$I then use the sum to infinity formula $S_{\infty} = \frac{a}{1 - r}$to get $\frac{4{\sqrt{n}} + 2n}{4 - n}$, add back in the 1 and simplify to get $\frac{\frac{4}{n} + \frac{4}{\sqrt{n}} + 1}{\frac{4}{n} - 1}$.

Finally, when I apply the limit of n to infinity, I get back -1. But, this does not seem right to me. Looking back at the equation, if n is positive, the sum to infinity should be a positive number instead.

My guess is that the ratio that I used when calculating sum to infinity is wrong. The ratio should be less than 1, but my ratio is more than 1 if n tends to infinity. How do I go about solving this?

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Let $n=k^2$ then $$\lim_{n \to \infty}(1 + \frac{1}{\sqrt{n}})^n=\lim_{k \to \infty}(1 + \frac{1}{k})^{k^2}=\lim_{k \to \infty}\left((1 + \frac{1}{k})^{k}\right)^k\to e^\infty=\infty$$

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Note that $\lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^{\sqrt n}=e>2$ so that for $n$ large enough we have $\left(1+\frac1{\sqrt n}\right)^{\sqrt n}>2$ and consequently: $$\left(1+\frac1{\sqrt n}\right)^{n}>2^{\sqrt n}$$

This shows that: $$\lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^{n}=+\infty$$

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  • $\begingroup$ I see... this proof is really clean and understandable, but are there any alternative approaches that make use of the binomial expansion of the equation? $\endgroup$ – statsguy21 Sep 4 '18 at 12:23
  • $\begingroup$ Can I simply use the formula of the sum to infinity that I derived and say that when n tends to $\infty$, ratio of $\frac{\sqrt{n}}{2}$ will be $\geq$ 1. Hence, the series will be divergent? $\endgroup$ – statsguy21 Sep 4 '18 at 12:35
  • $\begingroup$ With binomial expansion you can derive that $\cdots\geq1+\sqrt n$ which also proves divergence. $\endgroup$ – drhab Sep 4 '18 at 13:23
  • $\begingroup$ I see...seems like there's quite a lot of ways to prove this. Thanks! $\endgroup$ – statsguy21 Sep 5 '18 at 12:36
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Sep 5 '18 at 12:36
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By Bernoulli inequality $(1+x)^r\ge 1+rx$ we have

$$\left(1 + \frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\ge1 + \sqrt{n}\frac{1}{\sqrt{n}}=2$$

and therefore

$$\left(1 + \frac{1}{\sqrt{n}}\right)^{{n}}=\left[\left(1 + \frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right]^{\sqrt{n}}\ge2^{\sqrt{n}}\to \infty$$


As an alternative by Taylor's expansion we have

$$\left(1 + \frac{1}{\sqrt{n}}\right)^{n}=e^{n\log \left(1 + \frac{1}{\sqrt{n}}\right)}=e^{n\left(\frac{1}{\sqrt{n}}+O(1/n)\right)}=e^{\left(\sqrt n+O(1)\right)}\sim e^\sqrt n\to \infty$$

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You can use the binomial expansion of the equation:

$$\left(1 + \frac{1}{\sqrt{n}}\right)^n = \sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$

Since all terms are positives, you can fix a lower bound:

$$\sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} \ge \sum_{k=n-1}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$

Then note that:

$$\sum_{k=n-1}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} =$$$$ \begin{pmatrix} n\\n-1 \end{pmatrix} 1^{n-1}\times\left( \frac{1}{\sqrt{n}}\right)^1 + \begin{pmatrix} n\\n \end{pmatrix} 1^{n} \times\left(\frac{1}{\sqrt{n}}\right)^0 \\ $$$$=n\times\frac{1}{\sqrt{n}} + 1$$

So, you have

$$\left(1 + \frac{1}{\sqrt{n}}\right)^n \ge 1+\sqrt{n}$$

From there, you can deduce your limit!

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So, basically you are saying: $$(1 + \frac{1}{\sqrt{n}})^n\le \frac{\frac{4}{n} + \frac{4}{\sqrt{n}} + 1}{\frac{4}{n} - 1},$$ however, it is not true by WA for $n\ge 5$. It implies that your original estimate with the geometric progression is not valid.

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