Answer to Question #251516 in Real Analysis for Ichigo

Question

Check the convergence of
2n/3^(n+1) where n is from 1 to infinity

Answer 1

We can use D'alembert criterion.

Find the next limit: $lim$_n->∞${\frac {a{\scriptscriptstyle n+1}} {a{\scriptscriptstyle n}}}$, where $a\scriptscriptstyle n$ is the n-th term of the series.

If the value of this limit is less than 1, then the series is convergent. If it is greater than 1, then the series is not convergent. In case of it being equal to one, there is no possibility to define convergence or divergence (then a different method should be used).

In our case,

${\frac {a{\scriptscriptstyle n+1}} {a{\scriptscriptstyle n}}}={\frac {2(n+1)*3^{n+1}} {2n*3^{n+2}}} = {\frac {n+1} {3n}} = {\frac 1 3} + {\frac 1 {3n}}$

$lim{\scriptscriptstyle n\to {\infty}} {\frac {a{\scriptscriptstyle n+1}} {a{\scriptscriptstyle n}}}= {\frac 1 3} + lim{\scriptscriptstyle n\to {\infty}}{\frac 1 {3n}} = {\frac 1 3}$ < 1 $\to$ series is convergent