I don't want the full answer but hints for solving this question.My idea to attempt the question so far is the following by back tracing if we calculate the limit of $\dfrac{e^{a_n}}{n}$ we will get our answer just by applying logarithm.As $\dfrac{e^{a_n}}{n}$ might be useful in the sense that we can use the given conditions of the problem and with Taylor series expression.But how should I proceed to get to $\dfrac{e^{a_n}}{n}$. I think we need to construct a new sequence.Is my approach correct or I need to think differently? Hints required
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It should be exist – UmbQbify Jun 23 '20 at 19:35
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Hint: Apply the 'L' Hospital Rule' for sequences to get $\lim_{n \to \infty} \frac{e^{a_n}}{n} = \lim_{n \to \infty} (e^{a_{n+1}}-e^{a_n})$ – Subham Jaiswal Jun 23 '20 at 20:00
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1Since your sequence is monotone, you could try to estimate $\displaystyle\sum^n_{k=0}(a_{k+1}-a_k),e^{a_k}=n+1$ by an integral (from above and below). – Jun 23 '20 at 20:13
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2FYI: The source of this problem is 2012 Putnam Exam Problem B4. – JimmyK4542 Jun 23 '20 at 20:23
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Does this answer your question? limit of a recursively defined function – metamorphy Nov 29 '22 at 03:08
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Proposition 1. $$x+\frac{1}{e^x}\geq \log{\left(e^x+1\right)}, \forall x\geq0$$
Indeed, from $\forall x\geq 0$ $$\log{(1+x)}\leq x \Rightarrow \log{\left(1+\frac{1}{e^x}\right)}\leq \frac{1}{e^x} \Rightarrow \\ x+\log{\left(1+\frac{1}{e^x}\right)}\leq x+\frac{1}{e^x} \Rightarrow \\ \log{e^x}+\log{\left(1+\frac{1}{e^x}\right)}\leq x+\frac{1}{e^x} \Rightarrow \\ \log{\left(e^x+1\right)}\leq x+\frac{1}{e^x}$$
Proposition 2. $$a_n\geq\log{(n+1)},\forall n\geq0$$
By induction $$a_0=1>\log{(0+1)}=0$$ $$a_1=1+\frac{1}{e}>\log{(1+1)}=0.693...$$ Now, let's assume $a_n\geq\log{(n+1)}$, then $$a_{n+1}=a_n+\frac{1}{e^{a_n}} \overset{Prop. 1}{\geq} \log{(e^{a_n}+1)} \geq \log{(n+1+1)}=\log{(n+2)}$$
Proposition 3. $a_n -\log{n}$ is decreasing and bounded/positive, thus converging.
Indeed $$a_{n+1}-a_n=\frac{1}{e^{a_n}} \Rightarrow \\ \left(a_{n+1}-\log{(n+1)}\right)-\left(a_n-\log{(n+1)}\right)=\frac{1}{e^{a_n}} \Rightarrow\\ \left(a_{n+1}-\log{(n+1)}\right)-\left(a_n-\log{n}\right)= \frac{1}{e^{a_n}} -\log{\left(1+\frac{1}{n}\right)} \leq ...$$
we know that $$\frac{1}{n+1}\leq \log{\left(1+\frac{1}{n}\right)}$$ thus $$...\leq \frac{1}{e^{a_n}} -\frac{1}{n+1}\overset{Prop.2}{\leq}0$$ As a result $$\color{red}{0<} \log{(n+2)}-\log{(n+1)} \overset{Prop.2}{\leq} \color{red}{a_{n+1}-\log{(n+1)}\leq a_n-\log{n}}$$
The remaining part is to find the limit ...
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