What is the sum from n=1 to infinity of (3.5+1)^n ?

The sum from n=1 to infinity of (3.5+1)^n is diverges

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sum from n=1 to infinity of (3.5+1)^n

n = 1 \sum \infty (3.5 + 1)^{n}

diverges

Solution steps

n = 1 \sum \infty (3.5 + 1)^{n}

Simplify

= n = 1 \sum \infty 4. 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 4. 5^{n} - 4. 5^{0}

n = 0 \sum \infty 4. 5^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

What is the sum from n=1 to infinity of (3.5+1)^n ?
The sum from n=1 to infinity of (3.5+1)^n is diverges