Supposing that $0 \le b_n < 1$ $\forall n \in N_+$ and that $\lim_{n \to \infty} nb_n = 0$, how can I deduce that $\lim_{n \to \infty}(1+b_n)^n = 1$, I can prove that $\lim_{n \to \infty}(1-b_n)^n = 1$ but I am not sure if this helps or not.
Thanks!
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Take the log of the limit, then apply L'Hospital's rule? – Simply Beautiful Art Jan 09 '17 at 15:36
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@SimpleArt haven't used L'Hospital's yet, would rather do it using bernoulli's and comparisons or something – H. Wilde Jan 09 '17 at 15:36
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Do you know that $\log (1+x) \leqslant x$ for all $x > -1$? – Daniel Fischer Jan 09 '17 at 15:51
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@DanielFischer yeah, the answer below is the kind of thing I was looking to use though I think – H. Wilde Jan 09 '17 at 15:51
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1Then you know that $0 \leqslant n\log (1 + b_n) \leqslant nb_n$. – Daniel Fischer Jan 09 '17 at 15:52
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See theorem at the start of this answer http://math.stackexchange.com/a/1668179/72031 – Paramanand Singh Jan 09 '17 at 16:10
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Conditions on $b_n$ are unnecessary and we only need that $nb_n\to 0$. – Paramanand Singh Jan 09 '17 at 16:16
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Let $0 < \epsilon < 1$. Since $\lim\limits_{n\to \infty} nb_n = 0$, there exists a positive integer $N$ such that $b_n < \frac{\epsilon}{n}$ for all $n > N$. Thus, if $n > N$,
$$\lvert (1 + b_n)^n - 1\rvert = \sum_{k = 1}^n \binom{n}{k}b_n^k < \sum_{k = 1}^n \binom{n}{k}\frac{\epsilon^k}{n^k} \le \sum_{k = 1}^n \frac{\epsilon^k}{k!} < \sum_{k = 1}^n \frac{\epsilon}{2^{k-1}}< 2\epsilon$$
Consequently, $\lim\limits_{n\to \infty} (1 + b_n)^n = 1$.
kobe
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2This is (so far, one never knows what the future brings) the only correct answer here. – Daniel Fischer Jan 09 '17 at 16:07
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Will you please justify the two inequalities $\sum_{k=1}^N \binom{n}{k}\dfrac{\epsilon^k}{n^k}\le \sum_{k=1}^N \dfrac{\epsilon^k}{k!}$ – Learnmore Jan 09 '17 at 16:32
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@learnmore It follows from the fact that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$. Indeed, $$\binom{n}{k}\frac{1}{n^k} = \frac{n(n-1)\cdots(n-k+1)}{k!n^k} = \left(1 - \frac{1}{n}\right)\cdots\left(1 - \frac{k-1}{n}\right)\frac{1}{k!} \le \frac{1}{k!}$$ – kobe Jan 09 '17 at 16:38
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Hint for $(1+b_n)^n$.
By Binomial theorem, we get $$(1+b_n)^n=1+nb_n+\frac{n(n-1)}{2}{b_n}^2+\cdots {b_n}^n $$
So $$\lim_{n\rightarrow \infty}(1+b_n)^n = \lim_{n\to \infty} (1+nb_n+\frac{n(n-1)}{2}{b_n}^2+\cdots {b_n}^n )$$ All the terms go to $0$ because of the facts that $0\leq b_n<1$ and that $\lim_{n\to \infty} nb_n=0$.
Similarly for $(1-b_n)^n$.
MoebiusCorzer
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seeker
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2But the number of terms grows with $n$. One needs to show that the sum nevertheless tends to $0$. – Daniel Fischer Jan 09 '17 at 15:52