The series $\frac{x}{1.2}+\frac{x^{2}}{3.4}+\frac{x^{3}}{5.6}+.....,x0$ is:

Q: The series $\frac{x}{1.2}+\frac{x^{2}}{3.4}+\frac{x^{3}}{5.6}+.....,x>0$ is:

To determine the convergence of the series $\frac{x}{1.2} + \frac{x^2}{3.4} + \frac{x^3}{5.6} + \ldots$, we need to analyze the behavior of its terms as $n$ increases. The general term of the series can be expressed as: \[ a_n = \frac{x^n}{(2n-1)(2n)} \] ### Explanation for Option 1: Convergent if $x < 1$ The series resembles a power series of the form $\sum_{n=1}^{\infty} \frac{x^n}{n^p}$ with $p = 2$, but with a slightly different denominator. To determine convergence, we can apply the ratio test, which is suitable for series with terms involving powers of $x$. The ratio test states that a series $\sum a_n$ converges absolutely if: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \] For our series, the ratio of consecutive terms is: \[ \frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{(2n+1)(2n+2)} \cdot \frac{(2n-1)(2n)}{x^n} = x \cdot \frac{(2n-1)(2n)}{(2n+1)(2n+2)} \] Simplifying the expression, we find: \[ \frac{a_{n+1}}{a_n} \approx x \cdot \frac{1}{1} = x \quad \text{as } n \to \infty \] Thus, the limit is: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |x| \] According to the ratio test, the series converges if $|x| 0$, this condition simplifies to $x 1$** If $x > 1$, the ratio test gives $|x| > 1$, which implies divergence. Therefore, Option 2 is incorrect. - **Option 3: Divergent if $x < 1$** As shown above, the series converges when $x < 1$. Therefore, Option 3 is incorrect. - **Option 4: Divergent if $x = 1$** If $x = 1$, the series becomes $\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)}$, which is a convergent series by the comparison test with the $p$-series $\sum \frac{1}{n^2}$. Therefore, Option 4 is incorrect. In conclusion, the correct answer is Option 1: the series is convergent if $x < 1$.

Question

The series $\frac{x}{1.2}+\frac{x^{2}}{3.4}+\frac{x^{3}}{5.6}+.....,x>0$ is:

Accepted Answer

To determine the convergence of the series $\frac{x}{1.2} + \frac{x^2}{3.4} + \frac{x^3}{5.6} + \ldots$, we need to analyze the behavior of its terms as $n$ increases. The general term of the series can be expressed as: $$ a_n = \frac{x^n}{(2n-1)(2n)} $$ ### Explanation for Option 1: Convergent if $x < 1$ The series resembles a power series of the form $\sum_{n=1}^{\infty} \frac{x^n}{n^p}$ with $p = 2$, but with a slightly different denominator. To determine convergence, we can apply the ratio test, which is suitable for series with terms involving powers of $x$. The ratio test states that a series $\sum a_n$ converges absolutely if: $$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 $$ For our series, the ratio of consecutive terms is: $$ \frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{(2n+1)(2n+2)} \cdot \frac{(2n-1)(2n)}{x^n} = x \cdot \frac{(2n-1)(2n)}{(2n+1)(2n+2)} $$ Simplifying the expression, we find: $$ \frac{a_{n+1}}{a_n} \approx x \cdot \frac{1}{1} = x \quad \text{as } n \to \infty $$ Thus, the limit is: $$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |x| $$ According to the ratio test, the series converges if $|x| < 1$. Since $x > 0$, this condition simplifies to $x < 1$. Therefore, Option 1 is correct: the series is convergent if $x < 1$. ### Explanation for Other Options: - **Option 2: Convergent if $x > 1$** If $x > 1$, the ratio test gives $|x| > 1$, which implies divergence. Therefore, Option 2 is incorrect. - **Option 3: Divergent if $x < 1$** As shown above, the series converges when $x < 1$. Therefore, Option 3 is incorrect. - **Option 4: Divergent if $x = 1$** If $x = 1$, the series becomes $\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)}$, which is a convergent series by the comparison test with the $p$-series $\sum \frac{1}{n^2}$. Therefore, Option 4 is incorrect. In conclusion, the correct answer is Option 1: the series is convergent if $x < 1$.