{"question":{"question":{"content":"$\displaystyle\sum\limits_{n=1}^{\infty } \dfrac{(x+3)^n}{n} $ **What is the interval of convergence of the series?** [[☃ radio 1]]","images":{},"widgets":{"radio 1":{"type":"radio","alignment":"default","static":false,"graded":true,"options":{"choices":[{"content":"$-5 \le x<-1$","correct":false},{"correct":false,"content":"$-5 < x<-1$"},{"isNoneOfTheAbove":false,"correct":false,"content":"$-4 < x<-2$"},{"isNoneOfTheAbove":false,"correct":true,"content":"$-4 \le x<-2$"}],"randomize":true,"multipleSelect":false,"displayCount":null,"hasNoneOfTheAbove":false,"onePerLine":true,"deselectEnabled":false},"version":{"major":1,"minor":0}}}},"answerArea":{"calculator":false,"chi2Table":false,"periodicTable":false,"tTable":false,"zTable":false},"itemDataVersion":{"major":0,"minor":1},"hints":[{"replace":false,"content":"We use the ratio test. For $x\neq -3$, let $a_n= \dfrac{(x+3)^n}{n}$ . $\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| {x+3}\right| $ [[☃ explanation 1]] The series converges when $\left| {x+3}\right| <1$, which is when $-4

Interval of convergence

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