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Beliebt Rechnen >

sum from n=1 to infinity of n-2

Lösung

n = 1 \sum \infty n - 2

Lösung

divergiert

Schritte zur Lösung

n = 1 \sum \infty n - 2

Seriendivergenztest anwenden:

divergiert

= divergiert

Beliebte Beispiele

sum from n=3 to infinity of 6/((n^2-1))

n = 3 \sum \infty \frac{6}{( n ^{2} - 1 )}

sum from n=1 to infinity of n(1+n^2)^1

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

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

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

sum from n=1 to infinity of 8/(n^7+9)

n = 1 \sum \infty \frac{8}{n ^{7} + 9}

sum from n=1 to infinity of 5nsin(4/n)

n = 1 \sum \infty 5 n sin (\frac{4}{n})