What is the interval of convergence of #sum_1^oo (3x-2)^(n)/(1+n+n^(x) #?

Question

What is the interval of convergence of #sum_1^oo  (3x-2)^(n)/(1+n+n^(x) #?

Luke Alvarez · Accepted Answer

The series converges for #1/3<=x<1#

We can use the ratio test and see for what values of #x#:

#L=lim_(n->oo) |a_(n+1)/a_n| < 1#

Calculate: #L= lim_(n->oo) |(3x-2)^(n+1)/(1+n+1+(n+1)^x)(1+n+n^x)/(3x-2)^n|#

#L= lim_(n->oo) |(3x-2)^(n+1)/(3x-2)^n||(1+n+n^x)/(2+n+(n+1)^x)|#

#L= |(3x-2)|lim_(n->oo) |(1+n+n^x)/(2+n+(n+1)^x)|= |(3x-2)|#

Hence, the series will undoubtedly converge for:

#|(3x-2)| < 1#

that is intended for:

#1/3 < x < 1#

And certainly divergent for #x < 1/3" and "x>1#

On the boundaries, where #L=1# the test is in inconclusive, we must analyse the two cases:

For #x=1/3# the series becomes:

#sum_1^oo (-1)^n/(1+n+n^(1/3))#

If we format this as follows:

#-sum_1^oo (-1)^(n+1)/(1+n+n^(1/3))#

We can see that for every #n# we have that:

#(-1)^(n+1)/(1+n+n^(1/3)) < (-1)^(n+1)/n#

Additionally, the first series is convergent, just as the second series is the alternate armonic series.

For #x=1# the series becomes:

#sum_1^oo 1/(1+2n)#

The integral test can be used to demonstrate the non-convergence of this series:

#lim_(x->oo) int_1^x dt/(1+2t) = -ln3 + lim_(x->oo) 1/2ln(1+2x) = +oo#

In conclusion the series converges for #x in [1/3,1)#

Christopher Agresta · Answer

undefined To find the interval of convergence of the series $\sum_{n=1}^{\infty} \frac{(3x-2)^n}{1+n+n^x}$, we can use the ratio test. The ratio test states that a series $\sum_{n=1}^{\infty} a_n$ converges absolutely if $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$.

Applying the ratio test to the given series:

$$
\lim_{n \to \infty} \left| \frac{\frac{(3x-2)^{n+1}}{1+(n+1)+(n+1)^x}}{\frac{(3x-2)^n}{1+n+n^x}} \right| < 1
$$

After simplifying:

$$
\lim_{n \to \infty} \left| \frac{(3x-2)^{n+1}(1+n+n^x)}{(3x-2)^n(1+(n+1)+(n+1)^x)} \right| < 1
$$

$$
\lim_{n \to \infty} \left| \frac{(3x-2)(1+n+n^x)}{1+(n+1)+(n+1)^x} \right| < 1
$$

To find the interval of convergence, solve the inequality:

$$
|3x - 2| \lim_{n \to \infty} \frac{1+n+n^x}{1+n+ (n+1)^x} < 1
$$

By considering the behavior of the limit as $n$ approaches infinity, determine the range of $x$ for which the inequality holds true.