Ratio test with an integer power of an in numerator

[tmt1]

I have

$$\sum_{n = 1}^{\infty} \frac{2^n}{n^{100}}$$

and I need to find whether it converges or diverges.

I can use the ratio test to get:

$$\lim_{{n}\to{\infty}} \frac{2^{n + 1}\cdot n^{100}}{2^n \cdot (n + 1)^{100}}$$

But I'm not sure how to get the limit from this.

I know the limit of $\frac{n^{100}}{(n + 1)^{100}}$ would be $1$. But how would I get the limit of $\frac{2^{n + 1}}{2^n}$?

 

[topsquark]

I have

$$\sum_{n = 1}^{\infty} \frac{2^n}{n^{100}}$$

and I need to find whether it converges or diverges.

I can use the ratio test to get:

$$\lim_{{n}\to{\infty}} \frac{2^{n + 1}\cdot n^{100}}{2^n \cdot (n + 1)^{100}}$$

But I'm not sure how to get the limit from this.

I know the limit of $\frac{n^{100}}{(n + 1)^{100}}$ would be $1$. But how would I get the limit of $\frac{2^{n + 1}}{2^n}$?

You don't need to worry about it.

\(\displaystyle \frac{2^{n + 1}}{2^n} = 2\)

-Dan

 

[Prove It]

I have

$$\sum_{n = 1}^{\infty} \frac{2^n}{n^{100}}$$

and I need to find whether it converges or diverges.

I can use the ratio test to get:

$$\lim_{{n}\to{\infty}} \frac{2^{n + 1}\cdot n^{100}}{2^n \cdot (n + 1)^{100}}$$

But I'm not sure how to get the limit from this.

I know the limit of $\frac{n^{100}}{(n + 1)^{100}}$ would be $1$. But how would I get the limit of $\frac{2^{n + 1}}{2^n}$?

I'm hoping that your intuition at least told you that you should be expecting the series to be divergent, as the exponential part would be a divergent geometric series, and this part will end up much, much greater than the polynomial...

 

[tmt1]

I'm hoping that your intuition at least told you that you should be expecting the series to be divergent, as the exponential part would be a divergent geometric series, and this part will end up much, much greater than the polynomial...

I suppose $2^n$ will be much greater than $n^{100}$, and $\sum_{}^{} 2^n$, is of course divergent, as $2 > 1$. So, even though $\sum_{}^{} \frac{1}{n^{100}}$ is convergent, it would be outpaced by $\sum_{}^{} 2^n$. Is this what you mean?

 

[Prove It]

I suppose $2^n$ will be much greater than $n^{100}$, and $\sum_{}^{} 2^n$, is of course divergent, as $2 > 1$. So, even though $\sum_{}^{} \frac{1}{n^{100}}$ is convergent, it would be outpaced by $\sum_{}^{} 2^n$. Is this what you mean?

Exactly! :)