What is the sum from n=0 to infinity of i/(n^4) ?

The sum from n=0 to infinity of i/(n^4) is converges

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sum from n=0 to infinity of i/(n^4)

n = 0 \sum \infty \frac{i}{n ^{4}}

converges

Solution steps

i = 1 \sum \infty \frac{n}{i ^{4}}

Apply the constant multiplication rule:

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

= i \cdot n = 1 \sum \infty \frac{1}{n ^{4}}

Apply p-Series Test:

converges

= i converges

= converges

n = 0 \sum \infty \frac{n}{11 n ^{6} + 1}

n = 0 \sum \infty (0.25)^{n}

n = 0 \sum \infty \frac{n}{7 ^{n}}

n = 5 \sum \infty \frac{1}{( n - 2 ) ^{4}}

n = 2 \sum \infty 9 e^{- 2 n}

What is the sum from n=0 to infinity of i/(n^4) ?
The sum from n=0 to infinity of i/(n^4) is converges