What is the sum from n=1 to infinity of-2^n ?

The sum from n=1 to infinity of-2^n is diverges

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sum from n=1 to infinity of-2^n

n = 1 \sum \infty - 2^{n}

diverges

Solution steps

n = 1 \sum \infty - 2^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 1 \sum \infty 2^{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 2^{n} - 2^{0})

n = 0 \sum \infty 2^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{2}{( n + 2 ) n}

n = 5 \sum \infty \frac{4 ^{n}}{5 ^{n}}

n = 1 \sum \infty \frac{n + 6}{n ^{2}}

n = 1 \sum \infty \frac{2 n}{n ^{2} + 4}

k = 2 \sum \infty \frac{k}{ln ( k )}

What is the sum from n=1 to infinity of-2^n ?
The sum from n=1 to infinity of-2^n is diverges