The question is: test $$\sum_k ((e^{k^{1/4}}+11)^{0.5}-(e^{k^{1/4}}-11)^{0.5})^k$$ for convergence. I believe I have to use the root test to get rid of the exponent of $0.5$, but after that I am absolutely lost!
2026-04-04 12:07:12.1775304432
Please Help me with this complicated convergent test!
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I assume that you are looking at the series $$\sum_{k=N}^{\infty}\left(e^{k^{1/4}}+11\right)^{0.5}-\left(e^{k^{1/4}}-11\right)^{0.5}$$ where $N>(\log11)^{4}$ so that we are not taking the square root of a negative number. The hint that follows will also work in the case that you are only trying to prove that the terms go to zero.
Hint: Noticing that $$\left(e^{k^{1/4}}+11\right)^{0.5}-\left(e^{k^{1/4}}-11\right)^{0.5}=\frac{22}{\sqrt{e^{k^{1/4}}+11}+\sqrt{e^{k^{1/4}}-11}},$$ we may bound the summands above by $$\frac{22}{\sqrt{e^{k^{1/4}}+11}}\leq\frac{22}{\sqrt{e^{k^{1/4}}}}\leq22e^{-k^{1/4}}.$$ The series $\sum_{k=N}^{\infty}22e^{-k^{1/4}}$ converges by comparison to other well known series (I am leaving this part for you), and so we have proven that the original series converges.