$\infty\sum$ (i)

Calculus Level pending

$\small Given\ the\ infinite\ sum:$ $\small S= \frac{2^k}{1\cdot 2\cdot 3}+\frac{3^k}{2\cdot 3\cdot 4}+\frac{4^k}{3\cdot 4\cdot 5}+\cdots$

$\small For\ which\ k > 0\ values\ is\ this\ sum\ convergent\ or\ divergent,\ respectively?$

$\small a)\ k < 2\ \ and\ k\geq 2\\\small b)\ k < 2\ \ and\ k\geq 3\\\small c)\ k < 1\ \ and\ k\geq 3\\\small d)\ k < 0\ \ and\ k\geq 0\$

$\small Note: There\ is\ a\ criteria\ which\ allows\ you\ finish\ in\ less\ than\ a\ minute$

b d c a

1 solution

Tom Engelsman
Aug 2, 2020

The general term of this infinite series can be written as $\frac{(n+1)^{k-1}}{n(n+2)}$ . The series converges iff the degree of the denominator:

1) Is larger than the degree of the numerator,

2) Is larger than 1.

Otherwise, it diverges. For convergence, we require:

$2 - (k-1) > 1 \Rightarrow \boxed{k < 2}$ (i)

and for divergence:

$2-(k-1) \le 1 \Rightarrow \boxed{k \ge 2}$ (ii)

Hence, choice A is correct.