Note that since $u$ is a unit and

$ua = au, \tag 1$

we may write

$a = u^{-1}au, \tag 2$

and thus

$au^{-1} = u^{-1}a; \tag 3$

also, since $a$ is nilpotent there is some $0 < n \in \Bbb N$ such that

$a^n = 0, \tag 4$

and thus

$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$

we observe that

$u + a = u(1 + u^{-1}a), \tag 6$

and that, by virtue of (5),

$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$ $= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$

this shows that

$(1 + u^{-1}a)^{-1} = \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$

and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),

$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$

that is, $u + a$ is a unit.

Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$.

Note that since $u$ is a unit and

$ua = au, \tag 1$

we may write

$a = u^{-1}au, \tag 2$

and thus

$au^{-1} = u^{-1}a; \tag 3$

also, since $a$ is nilpotent there is some $0 < n \in \Bbb N$ such that

$a^n = 0, \tag 4$

and thus

$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$

we observe that

$u + a = u(1 + u^{-1}a), \tag 6$

and that, by virtue of (5),

$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$ $= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$

this shows that

$(1 + u^{-1}a)^{-1} = \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$

and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),

$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$

that is, $u + a$ is a unit.

Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.

Note that since $u$ is a unit and

$ua = au, \tag 1$

we may write

$a = u^{-1}au, \tag 2$

and thus

$au^{-1} = u^{-1}a; \tag 3$

also, since $a$ is nilpotent there is some $0 < n \in \Bbb N$ such that

$a^n = 0, \tag 4$

and thus

$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$

we observe that

$u + a = u(1 + u^{-1}a), \tag 6$

and that, by virtue of (5),

$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$ $= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$

this shows that

$(1 + u^{-1}a)^{-1} = \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$

and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),

$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$

that is, $u + a$ is a unit.

Nota Bene: The result proved above has an application to this question.

Note that since $u$ is a unit and

$ua = au, \tag 1$

we may write

$a = u^{-1}au, \tag 2$

and thus

$au^{-1} = u^{-1}a; \tag 3$

also, since $a$ is nilpotent there is some $0 < n \in \Bbb N$ such that

$a^n = 0, \tag 4$

and thus

$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$

we observe that

$u + a = u(1 + u^{-1}a), \tag 6$

and that, by virtue of (5),

$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$ $= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$

this shows that

$(1 + u^{-1}a)^{-1} = \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$

and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),

$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$

that is, $u + a$ is a unit.

Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$.