How can I tell if this series converges or not by the root test?
\[\sum_{k=1}^{\infty} (1+3/k)^{k^2}\]
Can I apply the root test twice, since if that limit converges, the other one will converge too?
remember that \[a _{k=1}/a _{k}\]
k+1****
Yeah, I guess for some reason I was thinking I needed to get it into a form without an exponent to see if it converged or not, but really all I have to do is evaluate the limit... right?
well it conv. \[\left| r \right|<1\]
The sum diverges. And now that I think of it, I think I was right.
Let me look into my Calc book.
That's correct, if the limit goes to infinity, the series diverges also.
you have to solve then take the lim if the answer is less then 1 then conv.
So yeah, you only need to apply it once.
Thanks guys, helped a bunch.
No problem :-)
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