Converging series

  • #1
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Mod note: Moved from Homework section
I know that ##1/n^4## converges because of comparison test with ##1/n^2## (larger series) converges .
how do I know ##1/n^2## converges?
coz I cannot compare it with ##1/n## harmonic series as it diverges.

@REVIANNA, if you post in the Homework & Coursework sections, you must include an attempt. Your question seemed like more of a general question, but not a homework problem, so I moved your thread.
 
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Answers and Replies

  • #2
Samy_A
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Hint: for ##n\geq 2##: ##\frac{1}{n²}\leq \frac{1}{n²-n}=\frac{1}{n(n-1)}##
 
  • #3
Ray Vickson
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Homework Statement


I know that ##1/n^4## converges because of comparison test with ##1/n^2## (larger series) converges .
how do I know ##1/n^2## converges?
coz I cannot compare it with ##1/n## harmonic series as it diverges.

What is "coz"?

Anyway, the slickest way to show that ##\sum 1/n^p ## converges if ##p > 1 ## and diverges if ##p = 1## (or ##p < 1##) is to note that for ##n \geq 2## we have
[tex] \int_n^{n+1} \frac{dx}{x^p} < \frac{1}{n^p} < \int_{n-1}^n \frac{dx}{x^p} [/tex]
so you can get easily-computed upper and lower bounds on ##\sum_{n=2}^N 1/n^p##.
 
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