BAB 6 : BARISAN DAN DERET

(a)

\[ p = \lim_{k\to+\infty} \frac{\frac{3^{k+1}}{(k+1)!}}{\frac{3^k}{k!}} = \lim_{k\to+\infty} \frac{3^{k+1}}{(k+1)!} \cdot \frac{k!}{3^k} = \lim_{k\to+\infty} \frac{3}{k+1} = 0 \]

6

Perhatikan bahwa \(\frac{3}{k+1} < 1\) jika \(k \ge 3\), yang berarti \(\sum_{k=3}^\infty \frac{3^k}{k!}\) konvergen. Sedangkan untuk \(\sum_{k=1}^2 \frac{3^k}{k!} = \frac{3}{1!} + \frac{3^2}{2!} = 3 + \frac{9}{2} = \frac{15}{2}\). Jadi, \(\sum_{k=1}^\infty \frac{3^k}{k!}\) konvergen.

(b)

\[ \begin{aligned} \rho &= \lim_{k\to+\infty} \left|\frac{\frac{(k+1)^2}{(k+1)^2+1}}{\frac{k^2}{k^2+1}}\right| \\ &= \lim_{k\to+\infty} \frac{(k+1)^2}{(k+1)^2+1} \cdot \frac{k^2+1}{k^2} \\ &= \lim_{k\to+\infty} \frac{(k^2+2k+1)(k^2+1)}{(k^2+2k+2)k^2} \\ % Simplified approach for limit, as per image (though image seems to skip steps) % For L'Hopital: lim (k^2 / (k^2+1)) = 1, so the limit of the ratio should be 1. % The image directly goes to infinity, implying a different calculation method for the denominator, or a typo. % I'll follow the image's result for the limit as +inf &= +\infty \end{aligned} \]

Karena \(\rho = +\infty\), dengan demikian \(\sum_{k=1}^\infty \frac{k^2}{k^2+1}\) divergen.

(a) Gunakan prinsip informal (I), sehingga bentuk \(\sum_{n=1}^\infty \frac{4}{3^{n+1}}\) menjadi \(\sum_{n=1}^\infty \frac{4}{3^n}\). Selanjutnya, perhatikan bahwa \(\sum_{n=1}^\infty \frac{4}{3^n}\) merupakan deret geometri dengan rasio \(r = \frac{1}{3}\). Karena \(|r| < 1\), akibatnya \(\sum_{n=1}^\infty \frac{4}{3^n}\) konvergen. Dengan demikian deret \(\sum_{n=1}^\infty \frac{4}{3^{n+1}}\) konvergen.

(b)

\[ \begin{aligned} \sum_{k=1}^\infty \left[ \frac{7}{3^k} + \frac{6}{(k+3)(k+4)} \right] &= \sum_{k=1}^\infty \frac{7}{3^k} + \sum_{k=1}^\infty \frac{6}{(k+3)(k+4)} \\ &= 7 \sum_{k=1}^\infty \frac{1}{3^k} + 6 \lim_{b\to\infty} \sum_{k=1}^b \frac{1}{(k+3)(k+4)} \\ &= 7 \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots \right] + 6 \lim_{b\to\infty} \sum_{k=1}^b \left[ \frac{1}{k+3} - \frac{1}{k+4} \right] \\ &= 7 \left( \frac{1/3}{1 - 1/3} \right) + 6 \lim_{b\to\infty} \left[ \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \dots + \left(\frac{1}{b+3} - \frac{1}{b+4}\right) \right] \\ &= 7 \left( \frac{1/3}{2/3} \right) + 6 \lim_{b\to\infty} \left[ \frac{1}{4} - \frac{1}{b+4} \right] \\ &= 7 \left( \frac{1}{2} \right) + 6 \left[ \frac{1}{4} - 0 \right] \\ &= \frac{7}{2} + \frac{6}{4} \\ &= \frac{7}{2} + \frac{3}{2} \\ &= \frac{10}{2} \\ &= 5 \end{aligned} \]

Misalkan bahwa \(f(x) = \frac{1}{x(\ln x)^p}\). Selanjutnya perhatikan bahwa

\[ \frac{d}{dx}[x(\ln x)^p] = (\ln x)^p + p(\ln x)^{p-1} = (\ln x)^{p-1} (\ln x + p), \]

7

karena \(p > 1\), akibatnya \((\ln x)^{p-1} (\ln x + p) > 0\), sehingga \(f(x)\) monoton turun. Selain itu, \(f(x)\) juga kontinu di \([2, \infty)\), sehingga dengan uji integral

\[ \begin{aligned} \int_{2}^{+\infty} \frac{1}{x(\ln x)^p} dx \end{aligned} \]

Misalkan \(u = \ln x\), sehingga \(du = \frac{1}{x} dx\).

\[ \begin{aligned} \int \frac{1}{x(\ln x)^p} dx &= \int \frac{1}{u^p} du \\ &= \frac{1}{-p+1} u^{-p+1} \\ &= \frac{1}{-p+1} (\ln x)^{1-p} \end{aligned} \]

\[ \begin{aligned} \int_{2}^{+\infty} \frac{1}{x(\ln x)^p} dx &= \lim_{b\to+\infty} \int_{2}^{b} \frac{1}{x(\ln x)^p} dx \\ &= \lim_{b\to+\infty} \left[ \frac{1}{-p+1} (\ln x)^{1-p} \right]_{2}^{b} \\ &= \lim_{b\to+\infty} \frac{1}{-p+1} \left[ (\ln b)^{1-p} - (\ln 2)^{1-p} \right] \end{aligned} \]

Karena \(p > 1\), akibatnya \(\lim_{b\to+\infty} (\ln b)^{1-p} = 0\), sehingga

\[ \begin{aligned} \int_{2}^{+\infty} \frac{1}{x(\ln x)^p} dx &= \frac{1}{-p+1} \left[ 0 - (\ln 2)^{1-p} \right] \\ &= \frac{1}{-p+1} \left[ -\frac{1}{(\ln 2)^{p-1}} \right] \\ &= \frac{1}{(p-1)(\ln 2)^{p-1}} \end{aligned} \]

Karena \(\int_{2}^{+\infty} \frac{1}{x(\ln x)^p} dx\) konvergen jika \(p > 1\), akibatnya \(\sum_{k=2}^\infty \frac{1}{k(\ln k)^p}\) konvergen jika \(p > 1\).

Misalkan \(a_k = \frac{k}{3^k}\), dan perhatikan bahwa

\[ \begin{aligned} a_k - a_{k+1} &= \frac{k}{3^k} - \frac{k+1}{3^{k+1}} \\ &= \frac{3k - (k+1)}{3^{k+1}} \\ &= \frac{2k - 1}{3^{k+1}} > 0 \end{aligned} \] sehingga \(a_k > a_{k+1}\),

Selanjutnya,

\[ \begin{aligned} \lim_{k\to+\infty} a_k &= \lim_{k\to+\infty} \frac{k}{3^k} \\ &= \lim_{k\to+\infty} \frac{1}{3^k \ln 3} \\ &= 0 \end{aligned} \]

Karena memenuhi \(a_1 > a_2 > a_3 > \dots\) dan \(\lim_{k\to+\infty} a_k = 0\), dengan demikian \(\sum_{k=1}^\infty \frac{k}{3^k}\) konvergen.