Show that the sequence $$a_n = \bigg(1 + \frac{1}{n}\bigg)^n$$ is bounded.
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Rebellos
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user619263
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3Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ? – Rebellos Nov 25 '18 at 10:22
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Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue? – Mikey Spivak Nov 25 '18 at 10:30
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Using the inequality $$ (1+a)^n\ge 1+na, \quad n\in\mathbb N, \,\,a\ge-1, $$ and the binomial theorem we obtain $$ 2=1+n\cdot\frac{1}{n}\le\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!n^k}=\sum_{k=0}^n\frac{n(n-1)\cdots(n-k+1)}{k!n^k}\\ \le \sum_{k=0}^n\frac{1}{k!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\le 2+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}}<3. $$
Yiorgos S. Smyrlis
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Hints:
$$\bigg(1 + \frac{1}{n}\bigg)^n = {n\choose 0}+{n \choose 1}\bigg(\frac{1}{n}\bigg)+{n \choose 2}\bigg(\frac{1}{n}\bigg)^2+{n \choose 3}\bigg(\frac{1}{n}\bigg)^3+... \tag{1}$$
$$\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...<\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\tag{2}$$
Try simplifying $(1)$ and letting $n \to \infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)
KM101
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