Determine whether the series is convergent or divergent 1/5+ 1/8+ 1/11+ 1/14+ 1/17+...
S sareen New member Joined Oct 9, 2009 Messages 24 Dec 2, 2009 #1 Determine whether the series is convergent or divergent 1/5+ 1/8+ 1/11+ 1/14+ 1/17+...
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Dec 3, 2009 #2 \(\displaystyle \sum_{n=1}^{\infty}{\frac{1}{3n+2} \ = \ \frac{1}{5}+\frac{1}{8}+\frac{1}{11}\ +\frac{1}{14} \ + \ \frac{1}{17} \ + ...\) The ratio test tells us nothing, limit = one, ergo, a different ploy must be utilize.\displaystyle The \ ratio \ test \ tells \ us \ nothing, \ limit \ = \ one, \ ergo, \ a \ different \ ploy \ must \ be \ utilize.The ratio test tells us nothing, limit = one, ergo, a different ploy must be utilize. Let f(x) = 13x+2, f′(x) = −3(3x+2)2 and x ≥ 1, hence a good candidate for the integral test.\displaystyle Let \ f(x) \ = \ \frac{1}{3x+2}, \ f'(x) \ = \ \frac{-3}{(3x+2)^{2}} \ and \ x \ \ge \ 1, \ hence \ a \ good \ candidate \ for \ the \ integral \ test.Let f(x) = 3x+21, f′(x) = (3x+2)2−3 and x ≥ 1, hence a good candidate for the integral test. Therefore, limb→∞∫1b13x+2 = limb→∞ln∣3x+2∣3]1b = ∞, hence the series diverges.\displaystyle Therefore, \ \lim_{b\to\infty}\int_{1}^{b}\frac{1}{3x+2} \ = \ \lim_{b\to\infty}\frac{ln|3x+2|}{3}\bigg]_{1}^{b} \ = \ \infty, \ hence\ the \ series \ diverges.Therefore, b→∞lim∫1b3x+21 = b→∞lim3ln∣3x+2∣]1b = ∞, hence the series diverges.
\(\displaystyle \sum_{n=1}^{\infty}{\frac{1}{3n+2} \ = \ \frac{1}{5}+\frac{1}{8}+\frac{1}{11}\ +\frac{1}{14} \ + \ \frac{1}{17} \ + ...\) The ratio test tells us nothing, limit = one, ergo, a different ploy must be utilize.\displaystyle The \ ratio \ test \ tells \ us \ nothing, \ limit \ = \ one, \ ergo, \ a \ different \ ploy \ must \ be \ utilize.The ratio test tells us nothing, limit = one, ergo, a different ploy must be utilize. Let f(x) = 13x+2, f′(x) = −3(3x+2)2 and x ≥ 1, hence a good candidate for the integral test.\displaystyle Let \ f(x) \ = \ \frac{1}{3x+2}, \ f'(x) \ = \ \frac{-3}{(3x+2)^{2}} \ and \ x \ \ge \ 1, \ hence \ a \ good \ candidate \ for \ the \ integral \ test.Let f(x) = 3x+21, f′(x) = (3x+2)2−3 and x ≥ 1, hence a good candidate for the integral test. Therefore, limb→∞∫1b13x+2 = limb→∞ln∣3x+2∣3]1b = ∞, hence the series diverges.\displaystyle Therefore, \ \lim_{b\to\infty}\int_{1}^{b}\frac{1}{3x+2} \ = \ \lim_{b\to\infty}\frac{ln|3x+2|}{3}\bigg]_{1}^{b} \ = \ \infty, \ hence\ the \ series \ diverges.Therefore, b→∞lim∫1b3x+21 = b→∞lim3ln∣3x+2∣]1b = ∞, hence the series diverges.
