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Let $ S_n $=$\int_0^1 \frac{nx^{n-1}}{1+x}dx $ for n$ \ge $1 then as n tends to infinity sequence tends to:

1.0 2.1 3. 1/2 4. Infinity

Is there any other way, than to first do integration, and then take limit?

Foggy
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  • http://math.stackexchange.com/questions/292251/limit-of-s-n-int-limits-01-fracnxn-11x-dx-as-n-to-infty?rq=1 – Alex Jan 27 '15 at 10:08
  • $$\require{cancel} S_n = \int_0^1 \frac{nx^{n-1}}{1+x}dx = \int_0^1 \frac{1}{1+x} dx^n = \color{red}{\cancelto{\frac12}{\color{gray}{\left[\frac{x^n}{1+x}\right]_0^1}}}

    \color{red}{\cancelto{0}{\color{gray}{\int_0^1 \frac{x^n}{(1+x)^2} dx}}} $$

     – achille hui Jan 27 '15 at 10:08
  • @achillehui: fantastic formatting, how do you get that? – Jack D'Aurizio Jan 27 '15 at 10:21
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    @JackD'Aurizio It uses the cancel extension of MathJAX. see this answer for some sample usages. – achille hui Jan 27 '15 at 12:02

1 Answers1

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As stated by achille hui in his pretty comment, integration by parts leads to: $$ S_n = \color{red}{\frac{1}{2}}+\int_{0}^{1}\frac{x^n}{(1+x)^2}\,dx $$ but obviously: $$ 0\leq \int_{0}^{1}\frac{x^n}{(1+x)^2}\,dx \leq \int_{0}^{1} x^n\,dx = \frac{1}{n+1}$$ so the limit is just $\displaystyle\color{red}{\frac{1}{2}}.$

Jack D'Aurizio
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