Examples from Tests for series convergence All examples from Tests for series convergence.
[ EXAMPLE ] For
∑ n = 2 ∞ 1 n ln n , \sum_{n=2}^{\infty}\frac1{n\ln n}, n = 2 ∑ ∞ n ln n 1 , the corresponding integral is
∫ 2 ∞ d x x ln x = lim b → ∞ ( ln ( ln b ) − ln ( ln 2 ) ) = ∞ . \int_2^\infty\frac{dx}{x\ln x}
=\lim_{b\to\infty}\bigl(\ln(\ln b)-\ln(\ln2)\bigr)
=\infty. ∫ 2 ∞ x ln x d x = b → ∞ lim ( ln ( ln b ) − ln ( ln 2 ) ) = ∞. Therefore the series diverges.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
∑ n = 1 ∞ n n 3 + 2 , \sum_{n=1}^{\infty}\frac n{n^3+2}, n = 1 ∑ ∞ n 3 + 2 n , we have
0 < n n 3 + 2 ≤ n n 3 = 1 n 2 . 0<\frac n{n^3+2}\le\frac n{n^3}=\frac1{n^2}. 0 < n 3 + 2 n ≤ n 3 n = n 2 1 . The p p p
For
∑ n = 2 ∞ 1 n 2 − 2 , \sum_{n=2}^{\infty}\frac1{\sqrt{n^2-2}}, n = 2 ∑ ∞ n 2 − 2 1 , we have n 2 − 2 < n \sqrt{n^2-2}<n n 2 − 2 < n
1 n 2 − 2 > 1 n . \frac1{\sqrt{n^2-2}}>\frac1n. n 2 − 2 1 > n 1 . The given series dominates the divergent harmonic series and therefore diverges.
For
∑ n = 1 ∞ 1 n 2 + 3 , \sum_{n=1}^{\infty}\frac1{\sqrt{n^2+3}}, n = 1 ∑ ∞ n 2 + 3 1 , the direct comparison with 1 / n 1/n 1/ n
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
( 2 n + 1 ) / ( n 3 − 2 ) 1 / n 2 ⟶ 2 , \frac{(2n+1)/(n^3-2)}{1/n^2}\longrightarrow2, 1/ n 2 ( 2 n + 1 ) / ( n 3 − 2 ) ⟶ 2 , the series
∑ 2 n + 1 n 3 − 2 \sum\frac{2n+1}{n^3-2} ∑ n 3 − 2 2 n + 1 converges like ∑ 1 / n 2 \sum1/n^2 ∑ 1/ n 2
For
( n − 1 ) / ( n 4 + 1 ) 1 / 3 1 / n 1 / 3 ⟶ 1 , \frac{(n-1)/(n^4+1)^{1/3}}{1/n^{1/3}}
\longrightarrow1, 1/ n 1/3 ( n − 1 ) / ( n 4 + 1 ) 1/3 ⟶ 1 , the series
∑ n − 1 ( n 4 + 1 ) 1 / 3 \sum\frac{n-1}{(n^4+1)^{1/3}} ∑ ( n 4 + 1 ) 1/3 n − 1 diverges like ∑ 1 / n 1 / 3 \sum1/n^{1/3} ∑ 1/ n 1/3
Also,
3 n 2 + 5 n 2 n ( n 2 + 1 ) 3 / 2 n ⟶ 1 , \frac{\dfrac{3n^2+5n}{2^n(n^2+1)}}{3/2^n}
\longrightarrow1, 3/ 2 n 2 n ( n 2 + 1 ) 3 n 2 + 5 n ⟶ 1 , so the series has the same convergence behaviour as a geometric series.
Finally, for
∑ n = 2 ∞ ln n n 3 + 1 , \sum_{n=2}^{\infty}\frac{\ln n}{\sqrt{n^3+1}}, n = 2 ∑ ∞ n 3 + 1 ln n , use ln n ≤ n 1 / 4 \ln n\le n^{1/4} ln n ≤ n 1/4 n n n
ln n n 3 + 1 ≤ n 1 / 4 n 3 / 2 = 1 n 5 / 4 . \frac{\ln n}{\sqrt{n^3+1}}
\le\frac{n^{1/4}}{n^{3/2}}
=\frac1{n^{5/4}}. n 3 + 1 ln n ≤ n 3/2 n 1/4 = n 5/4 1 . The series converges by direct comparison.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For ∑ n 2 / 2 n \sum n^2/2^n ∑ n 2 / 2 n
( n + 1 ) 2 / 2 n + 1 n 2 / 2 n = 1 2 ( 1 + 1 n ) 2 ⟶ 1 2 . \frac{(n+1)^2/2^{n+1}}{n^2/2^n}
=\frac12\left(1+\frac1n\right)^2
\longrightarrow\frac12. n 2 / 2 n ( n + 1 ) 2 / 2 n + 1 = 2 1 ( 1 + n 1 ) 2 ⟶ 2 1 . The series converges.
For ∑ 1 / n ! \sum1/n! ∑ 1/ n !
1 / ( n + 1 ) ! 1 / n ! = 1 n + 1 ⟶ 0 , \frac{1/(n+1)!}{1/n!}=\frac1{n+1}\longrightarrow0, 1/ n ! 1/ ( n + 1 )! = n + 1 1 ⟶ 0 , so the series converges.
For ∑ n ! / 2 n \sum n!/2^n ∑ n ! / 2 n
( n + 1 ) ! / 2 n + 1 n ! / 2 n = n + 1 2 ⟶ ∞ . \frac{(n+1)!/2^{n+1}}{n!/2^n}
=\frac{n+1}{2}\longrightarrow\infty. n ! / 2 n ( n + 1 )! / 2 n + 1 = 2 n + 1 ⟶ ∞. The terms do not approach zero, so the series diverges.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
a n = ( n n + 1 ) n 2 , a_n=\left(\frac n{n+1}\right)^{n^2}, a n = ( n + 1 n ) n 2 , we have
a n n = ( n n + 1 ) n = ( 1 + 1 n ) − n ⟶ e − 1 < 1. \sqrt[n]{a_n}
=\left(\frac n{n+1}\right)^n
=\left(1+\frac1n\right)^{-n}
\longrightarrow e^{-1}<1. n a n = ( n + 1 n ) n = ( 1 + n 1 ) − n ⟶ e − 1 < 1. Therefore ∑ a n \sum a_n ∑ a n
For
a n = ( arctan n ) n , a_n=(\arctan n)^n, a n = ( arctan n ) n , a n n = arctan n ⟶ π 2 > 1. \sqrt[n]{a_n}=\arctan n\longrightarrow\frac\pi2>1. n a n = arctan n ⟶ 2 π > 1. Therefore the series diverges.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] The positive magnitudes 1 / n 1/n 1/ n
∑ n = 1 ∞ ( − 1 ) n + 1 n \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n n = 1 ∑ ∞ n ( − 1 ) n + 1 converges by the alternating-series test. But its absolute series is the harmonic series,
which diverges. The alternating harmonic series is therefore conditionally convergent.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let
S = 1 − 1 2 + 1 3 − 1 4 + 1 5 − ⋯ . S=1-\frac12+\frac13-\frac14+\frac15-\cdots. S = 1 − 2 1 + 3 1 − 4 1 + 5 1 − ⋯ . Take one positive term followed by two negative terms:
1 − 1 2 − 1 4 + 1 3 − 1 6 − 1 8 + 1 5 − 1 10 − 1 12 + ⋯ . 1-\frac12-\frac14+\frac13-\frac16-\frac18
+\frac15-\frac1{10}-\frac1{12}+\cdots. 1 − 2 1 − 4 1 + 3 1 − 6 1 − 8 1 + 5 1 − 10 1 − 12 1 + ⋯ . Grouping in blocks gives
( 1 − 1 2 − 1 4 ) + ( 1 3 − 1 6 − 1 8 ) + ⋯ = ∑ k = 1 ∞ ( 1 2 k − 1 − 1 4 k − 2 − 1 4 k ) = S 2 . \begin{aligned}
&\left(1-\frac12-\frac14\right)
+\left(\frac13-\frac16-\frac18\right)+\cdots\\
&=\sum_{k=1}^{\infty}
\left(\frac1{2k-1}-\frac1{4k-2}-\frac1{4k}\right)
=\frac S2.
\end{aligned} ( 1 − 2 1 − 4 1 ) + ( 3 1 − 6 1 − 8 1 ) + ⋯ = k = 1 ∑ ∞ ( 2 k − 1 1 − 4 k − 2 1 − 4 k 1 ) = 2 S . The changed sum is a consequence of conditional convergence, not an algebraic contradiction.
[RIGHT CLICK] [ASK AI] Tests for series convergence Builds on Infinite series .