Vtyjkyuk

Decide whether the series are absolutely convergent, conditionally convergent, or divergent.

$$a) sum _n=1^infty (-1)^n frac(log n)^log nn^a , a>0$$

$$b) sum_n=1^infty frac (-1)^[log n]n$$ where [ ] is greatest integer symbol.

My attempt : the easiest way to think about this problem is the Leibniz test both $a)$ and $b)$ will be conditionally convergents.

Is it correct ?

any hints/solutions

thanks u

For the first one we have that

$$frac(log n)^log nn^ato infty$$

indeed by $x=e^y to infty$ as $yge e^2a$

$$frac(log x)^log xx^a=fracy^ye^aygefrace^2aye^ay=e^ayto infty$$

therefore the given series diverges.

For the second one we can write

$$sum_n=1^infty frac (-1)^[log n]n=sum_n=1^1 frac 1n-sum_n=2^2 frac 1n+sum_n=3^7 frac 1n-sum_n=8^20 frac 1n+ldots+(-1)^k+1sum_n=[e^k]^[e^k+1-1] frac 1n+ldots =sum_k=1^infty (-1)^k+1 a_k$$

that is an alternating sum of terms not convergent to zero and thus it diverges too, indeed as $kto infty$ by harmonic series

$$a_k=sum_n=[e^k]^[e^k+1-1] frac 1nsim log (e^k+1)-log(e^k)=k+1-k=1$$

For the second part note that for $n$ between $e^k$ and $e ^k+1$, $(-1)^[log n]$ has a constant sign. Compare $sum_[e^k]^[e^k+1] frac 1 n$ with the integral of $frac 1 x $ from $[e^k]$ to $[e^k+1]$ to see that the sum of the terms from $[e^k]$ to $[e^k+1]$ does not tend to $0$. Hence the series is divergent.