I am needing to figure out whether these problems are convergent or divergent can anyone help me?
T TYSONG New member Joined Jan 30, 2006 Messages 8 May 1, 2006 #1 I am needing to figure out whether these problems are convergent or divergent can anyone help me?
pka Elite Member Joined Jan 29, 2005 Messages 11,996 May 1, 2006 #2 Your first two depend on: p>1⇒limn→∞∫anduup=limn→∞[1(p−1)ap−1−1(p−1)np−1]=1(p−1)ap−1,(a>0)\displaystyle p > 1\quad \Rightarrow \quad \lim _{n \to \infty } \int\limits_a^n {\frac{{du}}{{u^p }}} = \lim _{n \to \infty } \left[ {\frac{1}{{\left( {p - 1} \right)a^{p - 1} }} - \frac{1}{{\left( {p - 1} \right)n^{p - 1} }}} \right] = \frac{1}{{\left( {p - 1} \right)a^{p - 1} }},\quad (a > 0)p>1⇒n→∞lima∫nupdu=n→∞lim[(p−1)ap−11−(p−1)np−11]=(p−1)ap−11,(a>0) The third depends upon knowing limn→∞e−n=0\displaystyle \lim _{n \to \infty } e^{ - n} = 0n→∞lime−n=0
Your first two depend on: p>1⇒limn→∞∫anduup=limn→∞[1(p−1)ap−1−1(p−1)np−1]=1(p−1)ap−1,(a>0)\displaystyle p > 1\quad \Rightarrow \quad \lim _{n \to \infty } \int\limits_a^n {\frac{{du}}{{u^p }}} = \lim _{n \to \infty } \left[ {\frac{1}{{\left( {p - 1} \right)a^{p - 1} }} - \frac{1}{{\left( {p - 1} \right)n^{p - 1} }}} \right] = \frac{1}{{\left( {p - 1} \right)a^{p - 1} }},\quad (a > 0)p>1⇒n→∞lima∫nupdu=n→∞lim[(p−1)ap−11−(p−1)np−11]=(p−1)ap−11,(a>0) The third depends upon knowing limn→∞e−n=0\displaystyle \lim _{n \to \infty } e^{ - n} = 0n→∞lime−n=0
