anonymous
  • anonymous
how to show the series diverges for the sum from k=2 to infinity k^(k-2)/3k
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{2}^{\infty} k ^{k-2}/3k\]
amistre64
  • amistre64
one thought is to do the integral test
anonymous
  • anonymous
but k on its power so i need to use ln?

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anonymous
  • anonymous
use the ratio test will be much easier
amistre64
  • amistre64
\[\sum_{2}^{\infty} k ^{k-\cancel{2}^3}/3\cancel{k}\]
anonymous
  • anonymous
i havent learned the ratio test yet =[
amistre64
  • amistre64
1/3 pulls out and your left with \[1/3\sum_{2}^{\infty} k ^{k-3}\]
amistre64
  • amistre64
this should be self evident that it blows up to infinity
anonymous
  • anonymous
it should be K^(k-2) not K^(K-2^3) tho
amistre64
  • amistre64
noone did k^k-2^3 thats a cancel and becomes a 3
amistre64
  • amistre64
we have 1/k = k^-1 k^k-2-1=k^k-3
anonymous
  • anonymous
should i use ln to (K-3) down first
amistre64
  • amistre64
dunno, integral test was my first thought but i abandoned it for something a bit logical :)
amistre64
  • amistre64
hitting post is like playing the lotto, will it freeze or post? lol
anonymous
  • anonymous
loll what do you mean
amistre64
  • amistre64
just trying to work the site. i go to hit post to respond and my system will freeze up sometimes
anonymous
  • anonymous
i don't have the problem, its probably because this site updates every second
amistre64
  • amistre64
but yeah, if you can show k^(k-3) does off into infinity then it diverges by definition
amistre64
  • amistre64
goes off into ...
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sum+k%5E%28k-3%29+from+2+to+inf

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