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)

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degeneratecircle eNotes educator| Certified Educator

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 :)

oldnick | Student


it's easy to see that: the serie:


verifies:  `0<=s(n)<=t(n)`


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))`


Using logatitms:

`(2n-3)log3>(5n-6)log2`  (1)

Averaging value of  `log3/log2 cong 3/2`  so  (1) becames:

`3(2n-3)>2(5n-6)`    `n<5`

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.