<s>[INST] What is the sum of first $n$ terms of the series $1+\frac{1}{4}+3+1+\frac{1}{4}+3+1+\frac{1}{4}+3+\ldots \ldots $ Question: What is  the sum of the first $n$ terms of the divergent series :
$$
1+\frac{1}{4}+3+1+\frac{1}{4}+3+1+\frac{1}{4}+3+\ldots \ldots
$$.
My trials: $(1+1+1+\dots)+(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\dots)+(3+3+3+\dots)$. $n$th term of fist, second, third series are respectively, $a_n=1, b_n=\frac{1}{4}, c_n=3$. What can I do now?
 [/INST] The series is divergent
$S_n=\frac {n}{3}\times \frac{17}{4}$, if $n=3k,k\in Z^+$
$S_n=\frac {n+1}{3}\times \frac{17}{4}-3$, if $n=3k-1,k\in Z^+$
$S_n=\frac {n+2}{3}\times \frac{17}{4}-\frac{13}{4}$, if $n=3k-2,k\in Z^+$
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