Given $$\varphi(x)= \sum_{n=0}^\infty e^{\frac{n}{\ln(x)}} $$
which converges by the geometric series test for $0<\Re(x)<1.$
Prove that $1-\frac{1}{\varphi(x)}=e^{\frac{1}{\ln(x)}}$
Not really sure how to show this. Any hints?
2026-03-26 23:04:43.1774566283
Question about a function defined by a series
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The series is geometric : $$ \varphi\left(x\right)=\sum_{n=0}^{+\infty}{\mathrm{e}^{\frac{n}{\ln{x}}}}=\sum_{n=0}^{+\infty}{\left(\mathrm{e}^{\frac{1}{\ln{x}}}\right)^{n}}=\frac{1}{1-\mathrm{e}^{\frac{1}{\ln{x}}}} $$