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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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    Jonas, you're right. I meant $I_n(p) = \int_1^n \frac{dx}{x^p}$.2011-02-14

1 Answers 1

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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 $1$.