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I am trying to prove that the sequence $I_n(p)=\int_1^n \frac{dx}{x^p}$ does or does not converge uniformly to the function $I(p)=\int_1^\infty \frac{dx}{x^p}$ for $p>1$.

It seems to me that $\sup |I_n(p)-I(p)|= \lim_{p \to 1} (I(p)-I_n(p))$. Is it enough, then, to show that for all $\epsilon>0$ there exists a natural number $N$ such that for $p\to 1^+$ and all $n\geq N$, $I(p)-I_n(p) < \epsilon$? Can I set $p = 1$?

Edit:
$I(p)-I_n(p) = \int_1^\infty \frac{dx}{x^p} - \int_1^n \frac{dx}{x^p} = \int_n^\infty \frac{dx}{x^p}$.

$\lim_{n\to\infty} \lim_{p\to 1} \int_n^\infty \frac{dx}{x^p} = \lim_{n\to\infty} \int_n^\infty \frac{dx}{x} = 0$

Is this correct?

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Do you mean $I_n(p)=\int_1^n\frac{dx}{x^p}$? I ask because with your definition $I_n(p)\to 0$ as $n\to\infty$ for all $p\gt1$. In any case, you can explicitly compute $I(p)-I_n(p)$. –  Jonas Meyer Feb 13 '11 at 21:11
    
It does not, $I_n \to 0$ and $I \neq 0$. –  Jonas Teuwen Feb 13 '11 at 22:45
    
@cardinal: (You may not be alerted of this, but maybe you'll read it anyway.) I rejected an edit of yours because in the preview it looked incorrect, but I think that I was mistaken and that it was the same edit Arturo made. Sorry about that. (On the other hand, the problem itself seems to have an error that still hasn't been clarified.) –  Jonas Meyer Feb 13 '11 at 23:56
    
Jonas, you're right. I meant $I_n(p) = \int_1^n \frac{dx}{x^p}$. –  please delete me Feb 14 '11 at 8:01

1 Answer 1

up vote 1 down vote accepted

The sequence doesn't converge uniformly -- $\lim_{p \to 1} (I(p)-I_n(p))$ doesn't exist, since

\[I(p)-I_n(p)=\frac{1}{(p-1)n^{p-1}}\;,\]

which diverges for $p\to 1^+$. That is, for given $n$ the difference is unbounded in $p$, and thus you can't find an $n_0$ such that the difference is bounded by $\epsilon$ for all $n>n_0$ and all $p$. This is because the integral diverges for $p=1$, and no matter how large you choose $n$ you're missing larger and larger contributions from the logarithmic tail as you take $p$ to $1$.

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