Question

(a). $\sum_{n=1}^{\infty} \frac{1}{4}$ (c). $\sum_{n=1}^{\infty} (\frac{-1}{3})^n$ (b). $\sum_{n=1}^{\infty} \frac{n^2+3}{3n^2-5}$ (d). $\sum_{n=1}^{\infty} \frac{1}{2^{n-2}}$

          (a). $\sum_{n=1}^{\infty} \frac{1}{4}$ 
(c). $\sum_{n=1}^{\infty} (\frac{-1}{3})^n$
(b). $\sum_{n=1}^{\infty} \frac{n^2+3}{3n^2-5}$
(d). $\sum_{n=1}^{\infty} \frac{1}{2^{n-2}}$

(a). ∑n=1^∞(1)/(4)
(c). ∑n=1^∞ ((-1)/(3))^n
(b). ∑n=1^∞(n^2+3)/(3n^2-5)
(d). ∑n=1^∞(1)/(2^n-2)

Added by Joan R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

08/30/2023

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n+3 = 1 + 3 = 4 Show more…

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n+3 (b). n=1 3n2-5 3n+1 dn=1 2n-2 c.=1

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Question

(a). $\sum_{n=1}^{\infty} \frac{1}{4}$ (c). $\sum_{n=1}^{\infty} (\frac{-1}{3})^n$ (b). $\sum_{n=1}^{\infty} \frac{n^2+3}{3n^2-5}$ (d). $\sum_{n=1}^{\infty} \frac{1}{2^{n-2}}$

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