I have a homework question that asks the following:
If $\sum_{n=0}^{∞} c_n*4^n$ is convergent, can we conclude that the following series is also convergent?
$\sum_{n=0}^{∞} c_n*(-4)^n$
Now my logic is that we achieve convergence when the series gets progressively smaller and smaller and eventually converges and doesn't go off "to infinity".
Since the negative sign makes this series smaller overall since many terms will be negative, why can't we necessarily conclude that this series will converge? It seems very intuitive to me and was my go-to solution for this question, but I got it wrong.
Thank you for eading.
First, one misconception about the alternating sign: $c_n$ could also have arbitrary, alternating, or whatever sign. So your argument fails here.
If the radius of convergence is exactly $4$, then all bets are off: convergence at $R$ does not imply convergence at $-R$ (and vice-versa). For instance, take $c_n = \frac{(-1)^n}{4^n n}$. You have $R=4$, and $$ \sum_{n=1}^\infty c_n 4^n = \sum_{n=1}^\infty \frac{(-1)^n}{n} $$ which converges, yet $$ \sum_{n=1}^\infty c_n (-4)^n = \sum_{n=1}^\infty \frac{1}{n} = \infty\,. $$
Note: If the radius had been $R>4$, then things would have been good.