What is the sum from n=1 to infinity}e^{-n of+1 ?

The sum from n=1 to infinity}e^{-n of+1 is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity}e^{-n of+1

n = 1 \sum \infty e^{- n} + 1

diverges

Solution steps

n = 1 \sum \infty e^{- n} + 1

Apply Series Divergence Test:

diverges

= diverges

\frac{d}{d x} (1 + \frac{1}{4 x})

\frac{1}{10} e^{4 t} + \frac{9}{10} e^{- 6 t}

f (x) = 2 x^{2} - 16 x + 35

\int (- 3 x + 4 x) d x

d y - (y - 7)^{2} d x = 0

What is the sum from n=1 to infinity}e^{-n of+1 ?
The sum from n=1 to infinity}e^{-n of+1 is diverges