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 show that the given series is convergent if -5 < x < 5 and for no other value of x, we can use the ratio test.
Let's denote the nth term of the series as aₙ = (n * x^(n-1))/(5^n).
The ratio test states that if the limit of the absolute value of the ratio of consecutive terms of a series exists and is less than 1, then the series converges.
Let's apply the ratio test:
Step 1: Find the ratio of consecutive terms.
R = |aₙ₊₁ / aₙ| = |((n+1) * x^n)/(5^(n+1))| / |(n * x^(n-1))/(5^n)|
Step 2: Simplify the ratio.
R = |((n+1) * x^n)/(5^(n+1))| * |(5^n)/(n * x^(n-1))|
= |(n+1) * x^n| / |n * x^(n-1)|
Step 3: Cancel out common terms.
R = (n+1)/n * |x|
Step 4: Take the limit of the ratio.
lim(n→∞) R = lim(n→∞) [(n+1)/n * |x|]
When -5 < x < 5, the term |x| is always less than 1.
Thus, as n approaches infinity, (n+1)/n approaches 1.
Therefore, lim(n→∞) R = |x|
Step 5: Analyze the limit.
For the series to converge, the limit of the ratio (|x|) must be less than 1.
If |x| < 1, then the series is convergent.
If |x| ≥ 1, then the series is divergent.
From the given range -5 < x < 5, we can see that |x| will always be less than 1. Therefore, the series converges if -5 < x < 5 and diverges for any other value of x.
Hence, we have shown 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.