The given series is $$\sum_{n=0}^{\infty} [2^n+(-1)^n]x^n$$

I applied ratio test to determine the radius of convergence (r.o.c).

Let $$a_n=[2^n+(-1)^n]$$. We denote, $$a_{n_1}=2^n, \space a_{n_2}=(-1)^n \\ \therefore \lim_{n \to \infty} |\frac{a_{n_1}}{a_{n_1+1}}|=\frac{1}{2}$$.

Similarly, $$\lim_{n \to \infty} |\frac{a_{n_2}}{a_{n_2+1}}|=(-1)$$.

Hence r.o.c is $|(-1)+\frac{1}{2}|=\frac{1}{2}$.
Is it a correct way to find r.o.c?

I have another doubt regarding the ratio test. Some texts say to use $\lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}|$ to find r.o.c and some texts suggest to use $\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}|$ to determine the r.o.c. Which one is correct to determine the r.o.c?

Any help or suggestion is highly appreciated.



$\displaystyle\sum_{n=0}^{\infty} [2^n+(-1)^n]x^n=\sum_{n=0}^{\infty}(2x)^n+\sum_{n=0}^{\infty}(-x)^n$

Now $\sum_{n=0}^{\infty}y^n$ will converge iff $|y|<1$

  • $\begingroup$ One fact I need to know to apply your hint . Between this two $\lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}|$ and $\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}|$, which one is correct to find r.o.c? $\endgroup$
    – vbm
    Mar 16 '18 at 9:38
  • $\begingroup$ @thevbm, Need to utilize the intersection of both $\endgroup$ Mar 16 '18 at 9:44

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