Use a Comparison Test (with geometric or p-series) to determine the convergence of the series: `sum_(n=1)^oo` 3^n/4^(n-3)
The Comparison Test is unnecessary here. Just note that
`(3^n)/(4^(n-3))=4^3((3^n)/(4^n))=64(3^n)/(4^n)=64(3/4)^n,` so the series is
`64(3/4)+64(3/4)^2+64(3/4)^3...,` which converges because it is a geometric series.
If you must use the Comparison Test, compare the series to itself :)
it's easy to see that: the serie:
Let we see where `t(n)<s(n),` we have:
`3^n/4^(n-3)>2^n /3^(n-3)` `3^(2n-3)>2^n 4^(n-3)` `3^(2n-3)>2^n 2^(2(n-3))`
Averaging value of `log3/log2 cong 3/2` so (1) becames:
wich means the serie `S(n)` is dominated by `T(n)` ,but a finite number of terms (first 5 terms),so by comparison test, if `T(n)` converges , `S(n)` does at the same.
Now we can write `T(n) ` as:
`T(n)=27sum_(n=1)^oo(2/3)^n=27lim_(n->oo) 2/3 (1-(2/3)^n)/(1-2/3)`
`=27 lim_(n->oo)2/3(1-(2/3)^n)/(1/3)=27lim_(n->oo)2(1-(2/3)^n)` `=54`
So `T(n)` converges, and `S(n)` too.