Find the value of $$\lim_{n\to \infty}\sum_{k=1}^n \frac{\binom{n}{k}}{n^k(k+3)}$$
I am not sure how to progress. I feel as though Squeeze theorem might somehow be used but I can't see to comprehend how.
Can someone please help?
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It definitely converges, since the expression within the limit is less than $\sum_{k=1}^n \binom{n}{k} \frac{1}{n^k} = \left(1+\frac{1}{n}\right)^n < e$. – Patrick Stevens May 30 '17 at 08:03
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Exactly, but The (k+3) term is throwing me off as to what the solution is. – Apocalyptic Warrior May 30 '17 at 08:38
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3A duplicate of Evalute $ \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)} $ – metamorphy Jun 19 '20 at 13:08
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Hint: $$\left(1+\frac{t}{n}\right)^n-1=\sum_{k=1}^n \binom{n}{k}\frac{t^k}{n^k}$$
and hence $$ \int_0^1 t^2\left(\left(1+\frac{t}{n}\right)^n-1\right)dt=\sum_{k=1}^n \binom{n}{k}\frac{1}{(k+3)n^k}$$
Now $(1+\frac{t}{n})^n\leq \exp(t)$, you can use the DCT.
Jack D'Aurizio
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Kelenner
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1(+1) Well done. So the limit is $\int_{0}^{1}t^2(e^t-1),dt = e-\frac{7}{3}$. – Jack D'Aurizio May 30 '17 at 13:46
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@Kelenner Can you please expand a little more on the second step? I haven't completely comprehended it. – Apocalyptic Warrior May 31 '17 at 08:44
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Put $f_n(t) t^2\left(\left(1+\frac{t}{n}\right)^n-1\right)$. Then we have that $f_n(t)\to t^2(\exp(t)-1)=g(t)$. We have also that $0\leq f_n(t)\leq g(t)$ as $(1+\frac{t}{n})^n\leq \exp(t)$. By the Dominated convergence theorem, $\int_0^1f_n(t)dt \to \int_0^1g(t)dt$. It remains to compute $\int_0^1g(t)dt$ (see the comment by Jack D'Aurizio) – Kelenner May 31 '17 at 09:26
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@Kelenner Hey, thanks for the reply.I understood this step. I actually was asking for the step by which you equated the integral and the summation. – Apocalyptic Warrior Jun 01 '17 at 08:50
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I probably do not understand your question. I have multiplied by $t^2$ and then integrated from $0$ to $1$ each term of the sum. – Kelenner Jun 01 '17 at 08:53