Finding if a series converges or diverges

[tmt1]

I have this series

$$\sum_{n = 1}^{\infty} \frac{\ln\left({n + 4}\right)}{{n}^{\frac{5}{2}}}$$

which I need to find whether it converges or diverges.

I can use the limit comparison test and set $a_n = \frac{\ln\left({n + 4}\right)}{{n}^{\frac{5}{2}}}$ and $b_n = \frac{1}{{n}^{\frac{5}{2}}}$

Thus I need to find the limit of $ \frac{{n}^{\frac{5}{2}} \ln\left({n + 4}\right)}{{n}^{\frac{5}{2}}}$ which simplifies to $\ln\left({n + 4}\right)$ which evaluates to $\infty$.

The limit or $L$ is both equal to infinity and greater than 0.

Also, $ \sum_{}^{} b_n$ converges since $p > 1$

but I'm not sure how to apply the rules of the comparison test to these results.

 

[Prove It]

I have this series

$$\sum_{n = 1}^{\infty} \frac{\ln\left({n + 4}\right)}{{n}^{\frac{5}{2}}}$$

which I need to find whether it converges or diverges.

I can use the limit comparison test and set $a_n = \frac{\ln\left({n + 4}\right)}{{n}^{\frac{5}{2}}}$ and $b_n = \frac{1}{{n}^{\frac{5}{2}}}$

Thus I need to find the limit of $ \frac{{n}^{\frac{5}{2}} \ln\left({n + 4}\right)}{{n}^{\frac{5}{2}}}$ which simplifies to $\ln\left({n + 4}\right)$ which evaluates to $\infty$.

The limit or $L$ is both equal to infinity and greater than 0.

Also, $ \sum_{}^{} b_n$ converges since $p > 1$

but I'm not sure how to apply the rules of the comparison test to these results.

I would say $\displaystyle \begin{align*} n \geq 2 \end{align*}$ we have $\displaystyle \begin{align*} \ln{ \left( n + 4 \right) } < n \end{align*}$, thus

$\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{\ln{ \left( n + 4 \right) }}{n^{\frac{5}{2}}} &< \sum_{n = 2}^{\infty} \frac{n}{n^{\frac{5}{2}}} \\ &= \sum_{n = 2}^{\infty} \frac{1}{n^{\frac{3}{2}}} \end{align*}$

Notice that $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{1}{n^{\frac{3}{2}}} \end{align*}$ is a convergent p-series, so that means $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{\ln{ \left( n + 4 \right) }}{n^{\frac{5}{2}}} \end{align*}$ is convergent by comparison. Adding any finite number of terms to a convergent infinite series will still be convergent, thus $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{\ln{ \left( n + 4 \right) }}{n^{\frac{5}{2}}} \end{align*}$ is also convergent.

 

[tmt1]

I would say $\displaystyle \begin{align*} n \geq 2 \end{align*}$ we have $\displaystyle \begin{align*} \ln{ \left( n + 4 \right) } < n \end{align*}$, thus

$\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{\ln{ \left( n + 4 \right) }}{n^{\frac{5}{2}}} &< \sum_{n = 2}^{\infty} \frac{n}{n^{\frac{5}{2}}} \\ &= \sum_{n = 2}^{\infty} \frac{1}{n^{\frac{3}{2}}} \end{align*}$

Notice that $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{1}{n^{\frac{3}{2}}} \end{align*}$ is a convergent p-series, so that means $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{\ln{ \left( n + 4 \right) }}{n^{\frac{5}{2}}} \end{align*}$ is convergent by comparison. Adding any finite number of terms to a convergent infinite series will still be convergent, thus $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{\ln{ \left( n + 4 \right) }}{n^{\frac{5}{2}}} \end{align*}$ is also convergent.

How do we show that $\displaystyle \begin{align*} \ln{ \left( n + 4 \right) } < n \end{align*}$ for $x \geq 2$?

 

[Prove It]

How do we show that $\displaystyle \begin{align*} \ln{ \left( n + 4 \right) } < n \end{align*}$ for $x \geq 2$?

Define $\displaystyle \begin{align*} f(x) = \ln{(x+4)} \end{align*}$ and $\displaystyle \begin{align*} g(x) = x \end{align*}$.

Notice that $\displaystyle \begin{align*} f(2) = \ln{(6)} \approx 1.792 \end{align*}$ and $\displaystyle \begin{align*} g(2) = 2 \end{align*}$. So $\displaystyle \begin{align*} f(2) < g(2) \end{align*}$.

Now we need to check that $\displaystyle \begin{align*} f(x) \end{align*}$ can never overtake $\displaystyle \begin{align*} g(x) \end{align*}$. To do this, we can check their derivatives.

$\displaystyle \begin{align*} f'(x) = \frac{1}{x + 4} \end{align*}$, so $\displaystyle \begin{align*} f'(2) = \frac{1}{6} \end{align*}$ and gets smaller as x gets larger. Meanwhile $\displaystyle \begin{align*} g'(x) = 1 \end{align*}$. So for all $\displaystyle \begin{align*} x \geq 2 \end{align*}$ we have $\displaystyle \begin{align*} 0 < f'(x) < g'(x) \end{align*}$. Since g(x) always grows at a faster rate than f(x), there is no way that f(x) can possibly overtake g(x). So that means $\displaystyle \begin{align*} \ln{(x + 4)} < x \end{align*}$ for all $\displaystyle \begin{align*} x \geq 2 \end{align*}$.

 

[alyafey22]

A standard approach is to set

$$F(x) = \ln(x+4)-x $$

We have

$$F'(x) = \frac{1}{x+4}-1 = \frac{-3-x}{x+4}$$

The derivative(slope) is clearly negative for $x\geq 2$ so $F$ is a decreasing function. Now since $F(2)<0$ we must have $F(x)<0, x\geq 2$. So $\ln(x+4)-x<0$ which implies that $\ln(x+4)<x$.