What is the sum from n=1 to infinity of (1.05)^n ?

The sum from n=1 to infinity of (1.05)^n is diverges

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sum from n=1 to infinity of (1.05)^n

n = 1 \sum \infty (1.05)^{n}

diverges

Solution steps

n = 1 \sum \infty 1.0 5^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty 1.0 5^{n} - 1.0 5^{0}

n = 0 \sum \infty 1.0 5^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{n}{4 n ^{3} + 3}

k = 2 \sum \infty (\frac{3}{5})^{2 k}

m = 2 \sum \infty \frac{4}{3 ^{m}}

n = 1 \sum \infty n e^{(- n^{2})}

n = 2 \sum \infty sin (n)

What is the sum from n=1 to infinity of (1.05)^n ?
The sum from n=1 to infinity of (1.05)^n is diverges