Answer on Question #50261 – Math – Complex Analysis
Decide whether the series is convergent or divergent
A. ∑ 8 n + i 2 − n 9 n \sum \frac{8^{n + i2^{-n}}}{9^n} ∑ 9 n 8 n + i 2 − n
B. ∑ ( n + i n + n i i n ) \sum \left(\frac{n + in + n^i}{i^n}\right) ∑ ( i n n + in + n i )
Solution
A) ∑ 8 n + i 2 − n 9 n = ( 8 9 ) n 8 i 2 − n ∑ 8 n + i 2 − n 9 n = ∑ ( 8 9 ) n 8 i / 2 n . \sum \frac{8^{n + i2^{-n}}}{9^n} = \left(\frac{8}{9}\right)^n 8^{i2^{-n}} \sum \frac{8^{n + i2^{-n}}}{9^n} = \sum \left(\frac{8}{9}\right)^n 8^{i/2^n}. ∑ 9 n 8 n + i 2 − n = ( 9 8 ) n 8 i 2 − n ∑ 9 n 8 n + i 2 − n = ∑ ( 9 8 ) n 8 i / 2 n .
By Cauchy criterion
q = lim n → ∞ ∣ ( 8 9 ) n 8 i / 2 n ∣ n = 8 9 < 1 , hence the series is convergent . q = \lim_{n \to \infty} \sqrt[n]{\left| \left(\frac{8}{9}\right)^n 8^{i/2^n} \right|} = \frac{8}{9} < 1, \text{ hence the series is convergent}. q = n → ∞ lim n ∣ ∣ ( 9 8 ) n 8 i / 2 n ∣ ∣ = 9 8 < 1 , hence the series is convergent .
B) ∑ ( n + i n + n i i n ) = ∑ ( n + i n + e i l n n i n ) \sum \left(\frac{n + in + n^i}{i^n}\right) = \sum \left(\frac{n + in + e^{ilnn}}{i^n}\right) ∑ ( i n n + in + n i ) = ∑ ( i n n + in + e i l nn ) . By Cauchy criterion
q = lim n → ∞ ∣ n + i n + e i l n n i n ∣ n = lim n → ∞ ∣ n + i n + e i l n n ∣ = lim n → ∞ ( n + cos ( l n n ) ) 2 + ( n + sin ( l n n ) ) 2 = + ∞ , hence \begin{aligned}
q &= \lim_{n \to \infty} \sqrt[n]{\left| \frac{n + in + e^{ilnn}}{i^n} \right|} = \lim_{n \to \infty} |n + in + e^{ilnn}| \\
&= \lim_{n \to \infty} \sqrt{(n + \cos(lnn))^2 + (n + \sin(lnn))^2} = +\infty, \text{ hence}
\end{aligned} q = n → ∞ lim n ∣ ∣ i n n + in + e i l nn ∣ ∣ = n → ∞ lim ∣ n + in + e i l nn ∣ = n → ∞ lim ( n + cos ( l nn ) ) 2 + ( n + sin ( l nn ) ) 2 = + ∞ , hence
the series is divergent.
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