I have the following sequence $$ y_n = \int_0^1 \frac{x^n}{x+5}\,dx, n = 0,1,\dots $$

So first question is, how to show that it's always positive?

Second is, how to show that it's monotonically decreasing. I think I have to prove something like this, but how?: $$ \int_0^1 \frac{x^{n+1}}{x+5}\,dx \leq \int_0^1 \frac{x^n}{x+5}\,dx $$

Third question is, how to prove that it converges to zero?


For your first question: integrals of positive functions are positive.

For your second: if $f \leq g$ then $\int f \leq \int g$. Note that $x^{n+1} \leq x^{n}$ on $[0,1]$ (note that this answers your first question too, take $f=0$).

For your third question: We have shown the sequence is monotonically decreasing, and it is positive so it is bounded below by 0. Thus it converges to its infimum, which corresponds to its limit. Note that $\frac{1}{x+5}$ is bounded on $[0,1]$ so you may say $$ \frac{x^{n}}{x+5} \leq C x^{n} $$ for some big constant $C>0$. Hence $$\int_0^1\frac{x^{n}}{x+5}dx \leq C\int_0^1 x^ndx = \frac{C}{n+1} \to 0$$ as $n \to \infty$. By the squeeze theorem it follows that $$\int_0^1 \frac{x^n}{x+5}dx \to 0.$$

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  • $\begingroup$ why is $x^{n+1} \leq x^n$ on [0,1] (do i take [0,1] because of the integral from 0 to 1? $\endgroup$ – mathNewbie Oct 23 '13 at 1:13
  • $\begingroup$ Consider $0 \leq x \leq 1$. Multiply this inequality by $x^n$, you find $0 \leq x^{n+1} \leq x^n$. Yes you take $[0,1]$ because this is the region over which we are integrating. $\endgroup$ – nullUser Oct 23 '13 at 1:15
  • $\begingroup$ Ah thanks, you are awesome =) Just one last question why is $c * \int_{0}^{1} x^n dx = \frac{C}{n+1}$ $\endgroup$ – mathNewbie Oct 23 '13 at 1:19
  • $\begingroup$ This is an elementary calculation. $\int x^n = \frac{x^{n+1}}{n+1}$ and apply the fundamental theorem of calculus. $\endgroup$ – nullUser Oct 23 '13 at 1:24
  • $\begingroup$ yeah i just got it, made a simple fault :D THANKS $\endgroup$ – mathNewbie Oct 23 '13 at 1:24

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