Answer :
Answer:
B.The series converges because r=1/4
Step-by-step explanation:
We are given that
[tex]\sum_{k=1}^{\infty}\frac{5k^2}{4^k}[/tex]
We have to find the correct option.
Ratio test :[tex]\lim_{k\rightarrow\infty} \mid \frac{a_{k+1}}{a_k}\mid =r[/tex]
If r< 1 then series convergent
If r>1 then the series divergent
If r=1 , test fails
[tex]r=lim_{k\rightarrow \infty}\mid\frac{\frac{5(k+1)^2}{4^{k+1}}}{\frac{5k^2}{4^k}}}\mid [/tex]
[tex]r=\lim_{k\rightarrow \infty}\mid{\frac{5(k+1)^2}{4^k\cdot 4}\times\frac{4^k}{5k^2}}\mid[/tex]
[tex]r=\lim_{k\rightarrow \infty}\mid{\frac{(k+1)^2}{4(k^2)}\mid[/tex]
[tex]r=\lim_{k\rightarrow \infty}\mid\frac{k^2(1+\frac{1}{k})^2}{4k^2}\mid[/tex]
[tex]r=\frac{1}{4}[/tex]
r<1
Therefore, the series converges .
B.The series converges because r=1/4