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sam30317
Does the series converge or diverge? Summation (n+3^n)/4^n n=0 to inf.
\[\sum_{n=0}^{\inf} \frac{n+3^n}{4^n}\]
I would seperate it in two parts: n/4^n+(3/4)^n Then I'd try the root test on both parts seperately.
So i would be left with (n/4) = infinity= div (3/4) <1 = conv.. Because one div, they both dierge???
only the 4 is raised to the power n.
in the first part I mean.
So you don't get n/4 when you do the root test.
\[\frac{\sqrt[n]{n}}{4}\]
lol, that looks more complicated. How would i solve that??
Well, \[lim_{n \rightarrow \infty} \sqrt[n]{n} =1\] It's standard thing, good idea to remember that limit.
So \[\frac{\sqrt[n]{n}}{4} \rightarrow \frac{1}{4}\] So that part does converge.
Oooo kaaay, lol. Thanks!! Im having a hard time remembering when all the tests conv or diverge. Like with n term test: if it equals 0, its inconclusive. And p series: p>1, conv. How did u manage to memorize them w/o confusing yourself??
I only know the ratio and the root test, not sure what n term test or p series is. For ratio and root test it's the same, <1 to converge, 1 inconlusive, so that's not too hard to remember.
I do know the n term test actually, thanks wiki.
Lol. I had a test on this stuff already and it was crazy how confused I got (i.e I failed it). Now my final is today and im going crazy, lol
Well the term test makes a lot of sense, if the limit doesn't equal zero, you're adding an infinite amount of numbers that aren't close to zero, so that's sure to diverge.
It's the first one to try, because it's easy to calculate.
Pseries, nth term, geom are easy to remember. I know about the root/ration , thanks to you! Now....on to others
general rule: if you see powers of n: use the ratio test, if you see factorials: ratio test.