What is the radius of convergence of #sum_1^oo x/n#?

Question

Christopher Allers · Accepted Answer

The series:

#sum_(n=0)^oo x^n/n#

is convergent for #x in [-1,1)# Applying the test of ratios: #lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) ( (x^(n+1)/(n+1)) / (x^n/n))# #lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) n/(n+1) abs(x^(n+1)/x^n)# #lim_(n->oo) abs(a_(n+1)/a_n) = absx lim_(n->oo) n/(n+1) =absx# So for #absx < 1# the series is absolutely convergent and for #absx > 1# the series is not convergent. For #absx =1# we have the two cases: #(1) " " x=-1#: The series turns into: #sum_(n=0)^oo (-1)^n/n# As: #lim_(n->oo) abs((-1)^n/n) = lim_(n->oo) 1/n =0# and #abs( (-1)^(n+1)/(n+1)) = 1/(n+1) < 1/n = abs((-1)^n/n)# The Leibniz theorem indicates that the series is convergent. #(2) " " x=1#: The series turns into: #sum_(n=0)^oo1/n# Take a look at the partial sum sequence: #s_n = sum_(k=1)^n 1/k# based on Cauchy's necessary condition this sequence is convergent only if for every #epsilon >0# we can find #N# such that for every #m,n >N#: #abs(s_n-s_m) < epsilon# Suppose #m>n# and #p=m-n#, then: #s_m = s_n + sum_(k=1)^p 1/(n+k)# and: #abs (s_n-s_m) = sum_(k=1)^p 1/(n+k)# Now, as for #k=1,...,p# we have that: #1/(n+k) >= 1/(n+p)# #abs (s_n-s_m) >= sum_(k=1)^p 1/(n+p)# that is: #abs (s_n-s_m) >= p/(n+p)# #abs (s_n-s_m) >= 1/(1+n/p)# So if we choose #epsilon < 1/2# and #m=2n# so that #p=n# we get: #abs (s_n-s_m) >= 1/2 > epsilon# which demonstrates that the series is not convergent since Cauchy's condition is not met.

Adam Adkins · Answer

undefined The radius of convergence of the series $\sum_{n=1}^{\infty} \frac{x}{n}$ can be found using the ratio test.

The ratio test states that for a series $\sum_{n=1}^{\infty} a_n$, if $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$, then:

1. If $L < 1$, the series converges absolutely.
2. If $L > 1$ or $L = \infty$, the series diverges.
3. If $L = 1$, the test is inconclusive.

For the series $\sum_{n=1}^{\infty} \frac{x}{n}$, $a_n = \frac{x}{n}$. Applying the ratio test:

$$
L = \lim_{n \to \infty} \left| \frac{\frac{x}{n+1}}{\frac{x}{n}} \right| = \lim_{n \to \infty} \left| \frac{x}{n+1} \cdot \frac{n}{x} \right| = \lim_{n \to \infty} \left| \frac{n}{n+1} \right| = 1
$$

Since $L = 1$, the ratio test is inconclusive.

To find the radius of convergence, we can consider the ratio test when $L = 1$, which is the boundary case. In this case, the series converges if $\left| \frac{x}{n+1} \cdot \frac{n}{x} \right| < 1$. Solving this inequality for $x$, we get:

$$
\left| \frac{x}{n+1} \cdot \frac{n}{x} \right| < 1 \implies \left| \frac{n}{n+1} \right| < 1 \implies \frac{n}{n+1} < 1 \implies n < n+1
$$

Since $n$ can be any positive integer, there is no restriction on $x$, meaning the series converges for all real numbers $x$.

Therefore, the radius of convergence of the series $\sum_{n=1}^{\infty} \frac{x}{n}$ is $\infty$.