Question

Let {a_n} be a sequence defined as a₁ =3, a_n+1 = (1/5)a_n ,show that {a_n } an converges to zero.

Answer 1

$Question:-\\ Let\space {a_n}\space be\space a\space sequence\space defined\space as\space a_1\space =3,\space a_{n+1}\space =\space (\frac{1}{5})a_n\space ,\\ show\space that\space {a_n\space }\space an\space converges\space to\space zero.\\ ------------------------------------- Solution:-\\ first\space of\space all,\space we\space find\space some\space terms\space of\space sequence\\ given\\ a_1\space =3,\space a_{n+1}\space =\space (\frac{1}{5})a_n\space ,\\ we\space have\space ;\\ a_1=3...(1)\\ for\space a_2\space ,\space put\space n=1\space \space in\space \space a_{n+1};\\ a_2=(\frac{1}{5})a_1\space \\ put\space a_1\space value\space from\space (1)\\ a_2=(\frac{1}{5})(3)\space \\ a_2=(\frac{3}{5})...(2)\space \\ for\space a_3\space ,\space put\space n=2\space \space in\space \space a_{n+1};\\ a_3=(\frac{1}{5})a_2\space \\ put\space a_2\space value\space from\space (2)\\ a_2=(\frac{1}{5})(\frac{3}{5})\space \\ a_2=(\frac{3}{5^2})...(3)\space \\ for\space a_4\space ,\space put\space n=3\space \space in\space \space a_{n+1};\\ a_4=(\frac{1}{5})a_3\space \\ put\space a_2\space value\space from\space (3)\\ a_4=(\frac{1}{5})(\frac{3}{5^2})\space \\ a_4=(\frac{3}{5^3})\space \\ similarly\space \\ a_{n-1}=(\frac{3}{5^{n-2}})\space \\ for\space a_n\space ,\space put\space n=(n-1)\space \space in\space \space a_{n+1};\\ a_n=(\frac{1}{5})a_{n-1}\space \\ put\space a_{n-1}\space \\ a_n=(\frac{1}{5})(\frac{3}{5^{n-2}})\space \\ a_n=(\frac{3}{5^{n-1}})\space \\ now\space we\space show\space that\space {a_n\space }\space an\space converges\space to\space zero.\\\space for\space check\space to\space converges\space ,\space we\space take\space limit\space of\space a_n\\ \lim\limits_{n\space \to\space \infin}\space a_n=\lim\limits_{n\space \to\space \infin}\space (\frac{3}{5^{n-1}})\space \\ \lim\limits_{n\space \to\space \infin}\space a_n=\space (\frac{3}{5^{\infin-1}})\space \\ \lim\limits_{n\space \to\space \infin}\space a_n=\space (\frac{3}{5^{\infin}})\space \\ \lim\limits_{n\space \to\space \infin}\space a_n=\space (\frac{3}{\infin})\space \\ \lim\limits_{n\space \to\space \infin}\space a_n=\space 0\space \\ hence\\ we\space say\space that,\space {a_n\space }\space an\space converges\space to\space zero.\\\space$