Does #a_n=(n + (n/2))^(1/n) # converge?

Question

Ezra Adams · Accepted Answer

The sequence is convergent.

#a_n=(n+n/2)^(1/n)=n^(1/n)(3/2)^(1/n)# so #lim_(n->oo)(n+n/2)^(1/n)=lim_(n->oo)n^(1/n)lim_(n->oo)(3/2)^(1/n)# but #lim_(n->oo)n^(1/n)=1# and #lim_(n->oo)(3/2)^(1/n)=1# so #lim_(n->oo)a_n=1# and the sequence converges.

Isabella Ackerman · Answer

undefined To determine if the sequence $a_n = \left(n + \frac{n}{2}\right)^{\frac{1}{n}}$ converges, we need to analyze its behavior as $n$ approaches infinity.

First, let's rewrite $a_n$ using properties of exponents:

$$a_n = \left(n \left(1 + \frac{1}{2}\right)\right)^{\frac{1}{n}} = \left(\frac{3}{2}n\right)^{\frac{1}{n}}$$

Now, we can use the limit definition of $e$ (the base of the natural logarithm) to evaluate the limit:

$$\lim_{n \to \infty} \left(\frac{3}{2}n\right)^{\frac{1}{n}} = e^{\lim_{n \to \infty} \frac{\ln\left(\frac{3}{2}n\right)}{n}}$$

By applying L'Hôpital's Rule to the indeterminate form $\frac{\infty}{\infty}$, we get:

$$= e^{\lim_{n \to \infty} \frac{\frac{d}{dn}\left(\ln\left(\frac{3}{2}n\right)\right)}{\frac{d}{dn}n}} = e^{\lim_{n \to \infty} \frac{\frac{\frac{3}{2}}{\frac{3}{2}n}}{1}}$$

$$= e^{\lim_{n \to \infty} \frac{1}{n}}$$

Since the denominator $n$ grows without bound as $n$ approaches infinity, the limit of $\frac{1}{n}$ approaches zero.

Thus, the limit of $a_n$ simplifies to $e^0 = 1$.

Therefore, the sequence $a_n = \left(n + \frac{n}{2}\right)^{\frac{1}{n}}$ converges, and its limit is 1.