a) a sris that onvrgs absolutly

b) a sris that onvrgs onitionally

onsir th statmnt " a sris that onvrgs absolutly "

onsir th sris $\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{k}{4^k}$

onsir

to hk whthr th sris $\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{k}{4^k}$onvrgs onitionally or absolutly , hk th onvrgn of th sris $\sum _{k=1}^{\infty }\left|{\left(-1\right)}^k\frac{k}{4^k}\right|$

if th sris $\sum _{k=1}^{\infty }\left|{\left(-1\right)}^k\frac{k}{4^k}\right|$ivrgs thn th sris $\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{k}{4^k}$is onvrgnt onitionally

an

if$\sum _{k=1}^{\infty }\left|{\left(-1\right)}^k\frac{k}{4^k}\right|$is onvrgnt thn th sris $\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k}$is absolutly onvrgnt

th ratio tst for absolut onvrgn $\sum _{k=1}^{\infty }{a}_k$b th sris with non-zro trms, if L=$\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|thn$

1.if L<1 sris onvrgs absolutly

2.if L>1 sris ivrgs

.if L=1 th tst is inonlusiv

alulat th valu of $\left|\frac{a_{k+1}}{a_k}\right|$

onsir th trm $a_k={\left(-1\right)}^k\frac{k}{4^k}$

by substituting k=k+1

$$\begin{gathered} a_{k+1}={\left(-1\right)}^{k+1}\frac{k+1}{4^{k+1}} \\ \left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{{\left(-1\right)}^{k+1}{\displaystyle \frac{k+1}{4^{k+1}}}}{{\left(-1\right)}^k{\displaystyle \frac{k}{4^k}}}\right| \\ =\left|\frac{{\displaystyle \frac{k+1}{4^k}}}{{\displaystyle \frac{k}{4^k}}}\right| \\ =\left|\frac{(k+1)4^k}{k{4}^{k+1}}\right| \\ =\left|\frac{(k+1)}{4k}\right| \end{gathered}$$

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{k+1}{4k}\right| \\ L=\frac{1}{4} \end{gathered}$$

sin l<1 th sris onvrgs

by ratio tst for absolut onvrgs th sris onvrgs absolutly

onsir th statmnt " a sris that onvrgs onitionally".

onsir th sris $\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{\sqrt{k(k+1)}}$

rwritting th sris $\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{\sqrt{k(1+k)}}=\sum _{k=1}^{\infty }{(-1)}^k{a}_k$

lt$\left\{a_k\right\}$b th squn of positiv numbrs .

if$a_{k+1}<a_{k}forvryk\ge 1an\underset{k\to \infty }{\lim }{a}_k=0$

th altrnating sris$\sum _{k=1}^{\infty }{(-1)}^{k+1}{a}_{k}an\sum _{k=1}^{\infty }{(-1)}^k{a}_k$both onvrg

th gnral trm of th sris $a_k=\frac{1}{\sqrt{k(k+1)}}$

by substituting k=k+1

$$\begin{gathered} a_{k+1}=\frac{1}{\sqrt{(k+1)(1+k+1)}} \\ =\frac{1}{\sqrt{(k+1)(k+2)}} \\ {a}_{k+1}<a_k \end{gathered}$$

th squn is monotoni rasing squn

th valu of $\underset{k\to \infty }{\lim }{a}_{k}is:$

$\underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\frac{1}{\sqrt{k(k+1)}}=0$

sin $\underset{k\to \infty }{\lim }{a}_k=0$

by th altrnating sris $\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{\sqrt{k(k+1)}}$onvrgs

to hk whthr th sris$\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{\sqrt{k(1+k)}}$onvrgs onitionally 0r absolutly. hk th onvrgn of th sris$\sum _{k=1}^{\infty }\left|{(-1)}^k\frac{1}{\sqrt{k(1+k)}}\right|$

if th sris $\sum _{k=1}^{\infty }\left|\frac{{(-1)}^{k+1}}{1+k^2}\right|$ivrgs thn th sris $\sum _{k=1}^{\infty }\frac{{(-1)}^{k+1}}{1+k^2}$is onitionally onvrgnt

if th sris $\sum _{k=1}^{\infty }\left|\frac{{(-1)}^{k+1}}{1+k^2}\right|$onvrgs thn th sris $\sum _{k=0}^{\infty }\frac{{(-1)}^{k+1}}{1+k^2}$is absolutlly onvrgnt

$\sum _{k=1}^{\infty }\left|{(-1)}^k\frac{1}{\sqrt{k(1+k)}}\right|=\sum _{k=1}^{\infty }{(-1)}^k\left|\frac{1}{\sqrt{k(1+k)}}\right|$

lt $\sum _{k=1}^{\infty }{a}_{k}an\sum _{k=1}^{\infty }{b}_k$b two sris with positiv trms

1. if $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=L$, whr L is any positiv ral numbr , thn ithr both sris onvrgs or both ivrgs

2.if $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=0an\sum _{k=1}^{\infty }{b}_k$onvrgs thn$\sum _{k=1}^{\infty }{a}_k$onvrgs

if $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\infty an\sum _{k=1}^{\infty }{b}_k$ivrgs thn$\sum _{k=1}^{\infty }{a}_k$ivrgs

onsir th sris $a_k=\frac{1}{\sqrt{k(1+k)}}an{b}_k=\frac{1}{k}$

th valu of $$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{1}{\sqrt{k(1+k)}}}}{{\displaystyle \frac{1}{k}}} \\ =\underset{k\to \infty }{\lim }\frac{k}{\sqrt{k(1+k)}} \\ =\underset{k\to \infty }{\lim }\frac{1}{\sqrt{(1+{\displaystyle \frac{1}{k}}})} \\ =0 \end{gathered}$$

now $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$is harmoni sris

hn, $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$is ivrgnt

th sris $\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{\sqrt{k(k+1)}}$onvrgs onitionally.