Ratio test to find radius of convergence: $\sum\limits_{n=0}^\infty (n+1)(t-1)^{2n}$ Use the ratio test on :
$$\sum\limits_{n=0}^\infty (n+1)(t-1)^{2n}$$
and find the radius of convergence.
I know to start the ratio test, we consider $\frac{a_{n+1}}{a_n}$
So I let:
$$a_n=(n+1)(t-1)^{2n}$$
$$a_{n+1}=(n+2)(t-1)^{2n+2}$$
Thus,
$$\frac{a_{n+1}}{a_n} = \frac{(n+2)(t-1)^{2n+2}}{(n+1)(t-1)^{2n}} = \frac{(n+2)(t-1)^{2n}(t-1)^2}{(n+1)(t-1)^{2n}} = \frac{(n+2)(t-1)^2}{(n+1)}$$
Then I take the limit:
$$(t-1)^2 \lim_{n\to\infty} \frac{n+2}{n+1}=(t-1)^2$$
Thus, for this series: 


*

*Convergence if $(t-1)^2<1$

*Divergence if $(t-1)^2>1$
So is the radius of convergence $1$?
 A: Your solution for the most part is correct, up until determining the radius of convergence. (Your solution is correct, but incomplete, and I suspect correct for the wrong reasons since you didn't explicitly solve for $t$.) We note: for convergence, we want
$$(t-1)^2 < 1$$
Taking the square root of both sides gives the following, depending on whether you take the negative or positive root:
$$t-1 < 1 \;\;\; \text{and} \;\;\; t-1 > -1$$
An equivalent formulation of this statement is $|t-1|<1$. (You can see why this statement holds easily by solving $x^2 < 1$. From this, solving for $t$:
$$t < 2 \;\;\; \text{and} \;\;\; t > 0$$
Thus, for all $t$ satisfying $0 < t < 2$, the series converges, and diverges otherwise. The width of this interval is $2$, thus giving a radius of convergence of $1$. If you wanted to go further into the details on the interval of convergence, you could also justify the non-inclusion of $t=0$ and $t=2$ fairly easily by plugging each into the original series, simplifying, and checking for convergence that way as well.
This would be starkly different if, say, we have $(3t-1)^2 < 1$ instead for this series, which would result in a radius of convergence of $1/3$. (I'll leave showing that as an exercise to the reader.) The key point here being to make sure your work is all justified and to see where it comes from. I could easily see a few number of ways you could just pick out $1$ as the radius of convergence from $(t-1)^2 < 1$ and, while having the right radius, you would be wrong if you applied it to similar problems.
In short, correct solution, just perhaps for not the right reasons, it's hard to say, and this thus should be an example of the importance of showing your work.
