**Problem 6:**Given the series:\[ S = \sum_{n=0}^{\infty} \frac{(-1)^n 2n}{3^n} = 0 - \frac{2}{3} + \frac{4}{9} - \frac{6}{27} + \frac{8}{81} + \ldots \]**Tasks:**(a) **Show steps and find** \( \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \).(b) **Based on (a), does \( S \) converge or diverge? Does** \[ T = \sum_{n=0}^{\infty} \frac{2n}{3^n} \] **converge or diverge?**

Given: S=∑n=0∞-1n 2n3n=0-23+49-627+881+...To Find: a To find limn→∞ an+1an.b Based on part a, to determine whether S converges or diverges.To determine whether T=∑n=0∞ 2n3n converges or diverges.Solution: (a) The series is given byS=∑n=0∞-1n 2n3n=0-23+49-627+881+...The nth term of this series is given byan=-1n 2n3n⇒an+1=-1n+1 2n+13n+1⇒an+1an=-1n+1 2n+13n+1-1n 2n3n=-1n+1 2n+13n+13n-1n 2n=-1n+13n=-n+13nSince, an+1an=-n+13n=n+13n, for n≥1Since, limn→∞an+1an=limn→∞n+13n=limn→∞n1+1n3n=limn→∞1+1n3=limn→∞1+1n3 Using limit chain rule=1+limn→∞1n3=1+03 Since, limn→∞1n=0=13Therefore, limn→∞an+1an=13From the result of part a, we have, limn→∞an+1an=13<1Since, limn→∞an+1an<1Hence, according to the Ratio Test, the series ∑n=0∞ -1n 2n3n, converges.Therefore, the series S converges.(b) Now, if the series is given byT=∑n=0∞ 2n3nHere, the nth term is given byan=2n3nan+1=2n+13n+1⇒an+1an=2n+13n+12n3n=2n+13n+13n2n=n+13n⇒an+1an=n+13n=n+13n, for n≥1Since, limn→∞an+1an=limn→∞n+13nSince, from part a, we have, limn→∞n+13n=13⇒limn→∞an+1an=13<1Since, limn→∞an+1an<1Hence, according to the Ratio Test, the series ∑n=0∞ an conveges.Therefore, the series T converges.