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a) $$\sum_{k=1}^\infty 3^{-2k+1}$$ b) $$\sum_{k=1}^\infty 3^{2k+1}$$

my Trial

a) $$\sum_{k=1}^\infty 3^{-2k+1} = \sum_{k=1}^\infty \frac{-3}{9^k}$$

I am blocked because I wanted to have it in the form $ar^n$ and later us

an4s
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Alio Badi
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  • $a=-3, r=1/9$ You've done it. – saulspatz Mar 25 '18 at 16:45
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    No, $$\sum_{k=1}^\infty 3^{-2k+1} = \sum_{k=1}^\infty \frac{-3}{9^k}$$ is incorrect. Should be $$\sum_{k=1}^\infty 3^{-2k+1} = \sum_{k=1}^\infty \frac{3}{9^k}$$ – GEdgar Mar 25 '18 at 16:52
  • These questions really boil down to pre calc exponent properties if you pretend the summation symbol isn’t there.. – Prince M Mar 25 '18 at 22:38
  • @AlioBadi Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 27 '18 at 14:14

2 Answers2

1

Note that

  • $\sum_{k=1}^\infty 3^{-2k+1}=3\sum_{k=1}^\infty \left(\frac19\right)^k$

and

  • $\sum_{k=1}^\infty 3^{2k+1}=3\sum_{k=1}^\infty 9^k$
user
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0

For a), $r=1/9$; for b), $r=9$. So b) diverges, but... well, I'll let you work out a).

J.G.
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