The series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$ is convergent if
(A) $x>e$
(B) $x
(C) $x<1/e$
(D) $x>1/e$
I think the answer is C, but I could not determine the condition... how to solve it. Plz help.
The series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$ is convergent if
(A) $x>e$
(B) $x
(C) $x<1/e$
(D) $x>1/e$
I think the answer is C, but I could not determine the condition... how to solve it. Plz help.
We have \begin{align*} x^{\sum_{i=1}^n \frac 1i} &\le x^{\log n + 1}\\ &= x \cdot \exp(\log n\cdot \log x)\\ &= x \cdot n^{\log x} \end{align*} and \begin{align*} x^{\sum_{i=1}^n \frac 1i} &\ge x^{\log n}\\ &= \exp(\log n\cdot \log x)\\ &= n^{\log x} \end{align*} and hence \[ \sum_{n=1}^\infty x^{\sum_{i=1}^n \frac 1i} < \infty \iff \sum_{n=1}^\infty n^{\log x} < \infty \] which is true exactly for $\log x < -1\iff x < \frac 1e$.