How do you use the limit comparison test on the series #sum_(n=1)^oo(n+1)/(n*sqrt(n))# ?

Question

Luke Alvarez · Accepted Answer

The series:

#sum_(n=1)^oo (n+1)/(nsqrtn)#

is divergent.

Note that: #(n+1)/(nsqrtn) = n/(nsqrtn) +1/(nsqrtn) = 1/sqrtn+1/(nsqrtn)# as for every #n in NN#: #1/(nsqrtn) > 0# it follows that: #(n+1)/(nsqrtn) > 1/sqrtn# and as the series: #sum_(n=1)^oo 1/sqrtn = sum_(n=1)^oo n^(-1/2)# is divergent based on the #p# series test, we can conclude that: #sum_(n=1)^oo (n+1)/(nsqrtn)# is also divergent.

Adam Adkins · Answer

undefined To use the limit comparison test on the series $ \sum_{n=1}^\infty \frac{n+1}{n\sqrt{n}} $, we compare it with a known convergent or divergent series.

Let's choose the series $ \sum_{n=1}^\infty \frac{1}{\sqrt{n}} $, which is a known convergent series.

Then, we take the limit as $ n $ approaches infinity of the ratio of the terms of the two series:

$$ \lim_{n \to \infty} \frac{\frac{n+1}{n\sqrt{n}}}{\frac{1}{\sqrt{n}}} $$

Simplify the expression:

$$ \lim_{n \to \infty} \frac{n+1}{n} = 1 $$

Since the limit is a finite positive number, and the series $ \sum_{n=1}^\infty \frac{1}{\sqrt{n}} $ is convergent, by the limit comparison test, the series $ \sum_{n=1}^\infty \frac{n+1}{n\sqrt{n}} $ is also convergent.

Luke Alvarez · Answer

undefined To use the limit comparison test on the series $\sum_{n=1}^{\infty} \frac{n+1}{n\sqrt{n}}$, you would choose another series whose convergence behavior is known and compare it to the given series.

In this case, we can consider the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$, which is a well-known series (the p-series with $p = \frac{1}{2}$).

Now, we take the limit as $n$ approaches infinity of the ratio of the terms of the two series:

$$\lim_{n \to \infty} \frac{(n+1)/(n\sqrt{n})}{1/\sqrt{n}}$$

Simplify this expression to find the limit:

$$\lim_{n \to \infty} \frac{n+1}{n\sqrt{n}} \times \frac{\sqrt{n}}{1}$$

$$= \lim_{n \to \infty} \frac{n+1}{n}$$

$$= 1$$

Since this limit is a finite, positive number, and the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ converges (as a p-series with $p = \frac{1}{2}$), then by the limit comparison test, the series $\sum_{n=1}^{\infty} \frac{n+1}{n\sqrt{n}}$ also converges.