Question

Which of the following series does not converge uniformly on \((0,1)\)?
(I) \(\displaystyle\sum_{n=1}^\infty \frac{x^n}{n+x}\)
(II) \(\displaystyle\sum_{n=1}^\infty \frac{x}{n^2+x}\)
(III) \(\displaystyle\sum_{n=1}^\infty \frac{x^n}{1+x^n}\)

• A. Only (I)
• B. Only (II)
• C. Only (I) and (III)
• D. Only (II) and (III)

Hint:

Use the Weierstrass M-test for series II; check the tail sup near x=1 for series I and III.

Answer 1

The Correct Option is C

Step 1: Recall the Weierstrass M-test: if \(|a_n(x)|\le M_n\) for all \(x\) in the domain and \(\sum M_n<\infty\), then \(\sum a_n(x)\) converges uniformly.

Step 2: Test series (II): \(\displaystyle\sum\frac{x}{n^2+x}\) on \((0,1)\). Since \(0<x<1\), \(\dfrac{x}{n^2+x}<\dfrac{x}{n^2}<\dfrac{1}{n^2}\) for every \(n\). As \(\sum\dfrac{1}{n^2}\) converges, the M-test gives uniform convergence for series (II).

Step 3: Test series (I): \(\displaystyle\sum\frac{x^n}{n+x}\) on \((0,1)\). Look at the tail \(R_N(x)=\sum_{n=N+1}^\infty\dfrac{x^n}{n+x}\). Each term is increasing in \(x\) on \((0,1)\), so as \(x\to1^-\), \[R_N(x)\to\sum_{n=N+1}^\infty\frac{1}{n+1}\] which is a tail of the divergent harmonic series, so it equals \(\infty\). Hence \(\sup_{x\in(0,1)}R_N(x)=\infty\) for every fixed \(N\), so the tail never becomes uniformly small. Series (I) does not converge uniformly.

Step 4: Test series (III): \(\displaystyle\sum\frac{x^n}{1+x^n}\) on \((0,1)\). For \(x\) close to \(1\), each term \(\dfrac{x^n}{1+x^n}\) is close to \(\dfrac12\) for every fixed \(n\) up to a large range, since \(x^n\) stays close to \(1\). So for the tail \(R_N(x)=\sum_{n=N+1}^\infty\dfrac{x^n}{1+x^n}\), choosing \(x\) close enough to \(1\) makes arbitrarily many consecutive terms close to \(\tfrac12\), making \(R_N(x)\) arbitrarily large. So \(\sup_{x\in(0,1)}R_N(x)=\infty\) for every \(N\), and series (III) also fails to converge uniformly.

Step 5: Only series (II) converges uniformly on \((0,1)\); series (I) and (III) do not.

\[\boxed{\text{Only (I) and (III)}}\]