What is the sum from n=1 to infinity of (1/3)^{k-n} ?

The sum from n=1 to infinity of (1/3)^{k-n} is diverges

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sum from n=1 to infinity of (1/3)^{k-n}

n = 1 \sum \infty (\frac{1}{3})^{k - n}

diverges

Solution steps

k = 1 \sum \infty (\frac{1}{3})^{n - k}

Apply Series Ratio Test:

diverges

= diverges

5 sec^{2} (x)

k = 1 \sum \infty \frac{x ^{k}}{k}

f (x) = sec (x) tan (x)

x^{2} - 2 x, 2 x - 3

\int_{0}^{1} \frac{93}{x ^{5}} d x

What is the sum from n=1 to infinity of (1/3)^{k-n} ?
The sum from n=1 to infinity of (1/3)^{k-n} is diverges