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Evaluate $\lim_{n\to\infty}\sum_{r=0}^n\frac{n\choose r}{(r+4)n^r}$

My Attempt:

I tried writing the terms,

$\frac{n\choose 0}{4n^0}+\frac{n\choose 1}{5n^1}+\frac{n\choose 2}{6n^2}+\frac{n\choose 3}{7n^3}+...$

but couldn't see what to do with it.

Then I multiplied and divided the general term with $(r+3)(r+2)(r+1)$

$\frac{(r+3)(r+2)(r+1)n!}{(r+4)(r+3)(r+2)r!(n-r)!n^r}$

In denominator, we'll get $(r+4)!$ but don't know what to do afterwards.

The answer given is $6-2e$.

aarbee
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  • There are a few duplicates out there, basically it consists in approximating $\binom{n}{r}$ by $\frac{n^r}{r!}$ so that the limit with the $n^r$ on denominator makes the $n$ disappear from the sum. – zwim Sep 09 '23 at 09:15

1 Answers1

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$\lim_{n\to\infty}\sum_{r=0}^n\frac{n\choose r}{(r+4)n^r}$=$\frac{n\choose 0}{4n^0}+\frac{n\choose 1}{5n^1}+\frac{n\choose 2}{6n^2}+\frac{n\choose 3}{7n^3}+...$

Now observe $\dfrac{1}{4}=\int_0^1(x^3)dx$, Similarly $\dfrac{1}{5}=\int_0^1(x^4)dx$

So,$\lim_{n\to\infty}$$\frac{n\choose 0}{4n^0}+\frac{n\choose 1}{5n^1}+\frac{n\choose 2}{6n^2}+\frac{n\choose 3}{7n^3}+...=\lim_{n\to \infty}\int_0 ^1 \frac{n\choose 0}{n^0} x^3dx+\int_0 ^1 \frac{n\choose 1}{n^1} x^4dx+.....$

$\implies \lim_{n\to\infty}\int_0 ^1 x^3dx\bigg(\frac{n\choose 0}{n^0}+\frac{n\choose 1}{n^1} x^+.....+\frac{n\choose n}{n^n} x^n\bigg)\implies\lim_{n\to\infty}\int_0^1x^3dx\bigg(1+\dfrac{x}{n}\bigg)^n$

$\implies \int_0^1x^3 e^x dx=((x^3-3x^2+6x-6)e^x)_0 ^1=6-2e$

Note: (1) $\lim_{n\to\infty}\bigg(1+\dfrac{x}{n}\bigg)^n=e^x$

Note: (2) $\int_0^1x^3 e^x dx=(x^3-3x^2+6x-6)e^x+c$ ,Using Integration By parts

mathophile
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