What is the sum from n=1 to infinity of 1/(4(n+1)) ?

The sum from n=1 to infinity of 1/(4(n+1)) is diverges

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sum from n=1 to infinity of 1/(4(n+1))

n = 1 \sum \infty \frac{1}{4 ( n + 1 )}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{4 ( n + 1 )}

Apply the constant multiplication rule:

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

= \frac{1}{4} \cdot n = 1 \sum \infty \frac{1}{n + 1}

Apply Series Limit Comparison Test:

diverges

= \frac{1}{4} diverges

= diverges

n = 4 \sum \infty \frac{2}{n ^{2} - n}

n = 0 \sum \infty ln (1 + n^{6})

n = 1 \sum \infty \frac{1}{10 ln ( n )}

n = 0 \sum \infty \frac{( - 5 ) ^{n}}{n}

n = 0 \sum \infty \frac{( 6 ^{n} )}{n !}

What is the sum from n=1 to infinity of 1/(4(n+1)) ?
The sum from n=1 to infinity of 1/(4(n+1)) is diverges