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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? – 2017-01-09
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    @SimpleArt haven't used L'Hospital's yet, would rather do it using bernoulli's and comparisons or something – 2017-01-09
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    Do you know that $\log (1+x) \leqslant x$ for all $x > -1$? – 2017-01-09
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    @DanielFischer yeah, the answer below is the kind of thing I was looking to use though I think – 2017-01-09
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    Then you know that $0 \leqslant n\log (1 + b_n) \leqslant nb_n$. – 2017-01-09
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    See theorem at the start of this answer http://math.stackexchange.com/a/1668179/72031 – 2017-01-09
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    Conditions on $b_n$ are unnecessary and we only need that $nb_n\to 0$. – 2017-01-09

2 Answers 2

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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, $\lim\limits_{n\to \infty} (1 + b_n)^n = 1$.

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    This is (so far, one never knows what the future brings) the only correct answer here. – 2017-01-09
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    Will you please justify the two inequalities $\sum_{k=1}^N \binom{n}{k}\dfrac{\epsilon^k}{n^k}\le \sum_{k=1}^N \dfrac{\epsilon^k}{k!}$ – 2017-01-09
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    @learnmore It follows from the fact that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$. Indeed, $$\binom{n}{k}\frac{1}{n^k} = \frac{n(n-1)\cdots(n-k+1)}{k!n^k} = \left(1 - \frac{1}{n}\right)\cdots\left(1 - \frac{k-1}{n}\right)\frac{1}{k!} \le \frac{1}{k!}$$ – 2017-01-09
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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 $(1-b_n)^n$.

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    But the number of terms grows with $n$. One needs to show that the sum nevertheless tends to $0$. – 2017-01-09