Let $$s_n=\int_{0}^{1}\frac{nx^{n-1}}{(1+x)}\,dx$$ for $n\geq 1$.
Then where sequence $(s_n)$ converges.
There are four options (A)$0$ (B)$\frac1{2}$ (C)$1$ (D)$\infty$
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asimath
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@PraphullaKoushik I have done $\int_{0}^{1}\frac{x^{n-1}}{1+x},dx=\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-... $. Then I cannot do anything. – asimath Jan 15 '14 at 17:12
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Do you understand the hint given below? – Jan 15 '14 at 17:17
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@PraphullaKoushik Yes i think limit is 1/2. sorry for previous one – asimath Jan 15 '14 at 17:22
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yes... that is it! – Jan 15 '14 at 17:23
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Hmm, I keep getting $s_n=n \cdot \sum_{k=1}^{n-1}\frac{(-1)^{k+1}}{k}$ which diverges. Where did I make a mistake? – Alex Jan 15 '14 at 17:52
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1Does this answer your question? Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ – StubbornAtom Jan 27 '20 at 18:46
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HINT:
$$\frac{s_n}n+\frac{s_{n-1}}{n-1}=?$$
lab bhattacharjee
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I do not understand how does that help...$$\frac{s_n}n+\frac{s_{n-1}}{n-1}=\int_{0}^{1}\frac{x^{n-1}}{(1+x)},dx+ \int_{0}^{1}\frac{x^{n-2}}{(1+x)},dx=\int_{0}^{1}\frac{x^{n-1}+x^{n-2}}{(1+x)} ,dx$$ but then.. :O – Jan 15 '14 at 17:08
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@mathemagician : Oh yeah!!... $$\frac{s_n}n+\frac{s_{n-1}}{n-1}=\int_{0}^{1}\frac{x^{n-1}}{(1+x)},dx+ \int_{0}^{1}\frac{x^{n-2}}{(1+x)},dx=\int_{0}^{1}\frac{x^{n-1}+x^{n-2}}{(1+x)} ,dx=\int_{0}^{1}x^{n-2},dx= :) :)$$ Done... – Jan 15 '14 at 17:17
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@rajkamalmath, We have $$s_n+\frac{s_{n-1}}{\frac1{1-\frac1n}}=\frac1{\frac1{1-\frac1n}}$$
$$\lim_{n\to\infty}s_n+\frac{\lim_{n\to\infty}s_{n-1}}{\lim_{n\to\infty}\frac1{1-\frac1n}}=\frac1{\lim_{n\to\infty}\frac1{1-\frac1n}}$$
$$\lim_{n\to\infty}s_n+\lim_{n\to\infty}s_{n-1}=1$$ If $$\lim_{n\to\infty}s_n=S,\lim_{n\to\infty}s_{n-1}=? $$
– lab bhattacharjee Jan 15 '14 at 18:33 -
@labbhattacharjee : so S=$\frac{1}{2}$, for that reason I have told limit is $\frac1{2}$. Thanks! – asimath Jan 15 '14 at 20:25