What is the Convergence of the Integral for this Sequence?

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SUMMARY

The discussion centers on the convergence of the integral for the sequence defined by the function f_{n}(x)=n^{\alpha} for |x|≤1/n and f_{n}=0 elsewhere. The integral evaluates to 2n^{\alpha-1}, leading to the conclusion that the limit as n approaches infinity diverges to +∞ if α > 1. The user initially questioned the validity of changing the integral limits, but the correct evaluation confirms the divergence condition based on the value of α.

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Theraven1982
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Hello, it's me again ;)

Problem:
-------
Define

[tex]f_{n}(x)=n^{\alpha}, |x|\leq 1/n, f_{n}=0[/tex] elsewhere

Give all [tex]\alpha \in \Re[/tex] for which

[tex]\lim_{n \to \infty} \int_{\Re}f_{n}(x)dx=+\infty[/tex]
-------


Can i change this last integral to:

[tex]\lim_{x \to 0} \int_{0}^{\infty} x^{-\alpha}dx=+\infty[/tex]

But i think the integration limits aren't correct, and therefore [tex]alpha[/tex] is wrong too.

Any help appreciated,

kind regards,

W.
 
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Theraven1982 said:
Hello, it's me again ;)

Problem:
-------
Define

[tex]f_{n}(x)=n^{\alpha}, |x|\leq 1/n, f_{n}=0[/tex] elsewhere

Give all [tex]\alpha \in \Re[/tex] for which

[tex]\lim_{n \to \infty} \int_{\Re}f_{n}(x)dx=+\infty[/tex]

Let's see, your function f_n is a constant n^a on the interval [-1/n, 1/n], and zero elsewhere. So the integral is just

[tex]\int_{\Re}f_{n}(x)dx= n^a \cdot \frac{2}{n}= 2 n^{a-1}[/tex]

as you can easily see from the graph of f_n.

You can take it from there...
 
Off course... now that I see it, it's all very simple. Guess sometimes my mind gets confused after too much maths ;).
Thank you,

W.
 

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