Show that the series 1 + 2x/5 + 3x^2/25 + 4x^3/125 + ....is convergent if -5<x<5 and for no other value of x.
To determine the convergence of the series, we need to find the range of values for x where the series converges. The given series is:
1 + (2x/5) + (3x^2/25) + (4x^3/125) + ...
We can rewrite the series in sigma notation as:
∑[n=1 to ∞] (nx^(n-1)/(5^n))
Let's use the ratio test to analyze the convergence of the series. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges.
Let's compute the limit of the ratio of consecutive terms:
lim[n→∞] |(n+1)x^n/(5^(n+1))| / |nx^(n-1)/(5^n)|
Simplifying, we get:
lim[n→∞] |(n+1)x^n/(5^(n+1))| * |5^n/(nx^(n-1))|
= |(n+1)x^n/(5^(n+1)) * 5^n/(nx^(n-1))|
= |(n+1)x^n * 5^n/(5^(n+1) * nx^(n-1))|
= |(n+1)x^n * (5^n/(5*nx^(n-1)))|
= |(n+1)/5nx|
Taking the limit as n approaches infinity, we get:
lim[n→∞] |(n+1)/5nx|
Since x is a constant, the numerator (n+1) and denominator (5n) both grow linearly with n. Therefore, as n approaches infinity, the limit of the ratio is:
lim[n→∞] |(n+1)/5nx| = ∞/∞
To resolve the indeterminate form, we can apply L'Hôpital's rule. Taking the derivative of the numerator and denominator with respect to n:
d/dn (n+1) = 1
d/dn (5n) = 5
The derivative of the ratio becomes:
lim[n→∞] |1/5nx| = 1/5x * lim[n→∞] |1/n|
As n approaches infinity, 1/n approaches 0. Therefore, the limit of the ratio simplifies to:
lim[n→∞] |1/5nx| = 0
Since the limit is less than 1, the series converges. However, we need to consider the range of x for which this convergence holds.
In the derivation, we used the fact that x is a constant (|x| < 5). Therefore, the series converges for values of x where |x| < 5. In other words, the series converges if -5 < x < 5.
To summarize, the given series 1 + (2x/5) + (3x^2/25) + (4x^3/125) + ... is convergent if -5 < x < 5 and diverges for any other value of x.